21. Properties of Triangles’ Solutions Part 4#

The diagram is given below:

Let $ABC$ be the triangle. Let $O$ be the circumcenter and $I,$ the incenter.

Clearly, $OA = OB = OC = R, IE=r[IE\perp AB]$

Let $OM\perp BC$ then $\angle BOM = \angle COM = A$

Now, $OA = R, AI = r\cosec\frac{A}{2} = \frac{4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{\sin\frac{A}{2}}$

$= 4R\sin\frac{B}{2}\sin\frac{C}{2}$

$\angle OAB = \angle OBA = B - (90^\circ - A) = A + B - 90^\circ = 90^\circ - C$

$\therefore \angle OAI = \angle BAI - \angle BAO = \frac{A}{2} - (90^\circ - C)$

$= \frac{C - B}{2}$

Applying cosine law in $\triangle OAI,$

$\cos\frac{C - B}{2} = \frac{OA^2 + AI^2 - OI^2}{2OA.AI}$

$= \frac{R^2 + 16R^2\sin^2\frac{B}{2}\sin^2\frac{C}{2} - OI^2}{2.R.4R\sin\frac{B}{2}\sin\frac{C}{2}}$

$OI^2 = R^2\left[1 + 8\sin\frac{B}{2}\sin\frac{C}{2}\left\{2\sin\frac{B}{2}\sin\frac{C}{2} - \cos\left(\frac{B - C}{2}\right)\right\}\right]$

$= R^2\left[1 + 8\sin\frac{B}{2}\sin\frac{C}{2}\left\{\cos\frac{B - C}{2} - \cos\frac{B + C}{2} - \cos\frac{B - C}{2}\right\}\right]$

$= R^2\left[1 - 8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right]$

$= R^2 - 2Rr$

Let us prove the necessary condition for the second part.

Let $b$ be the A.M. of $a$ and $c$ i.e. $2b = a + c$

$\Rightarrow 2\sin B = \sin A + \sin C \Rightarrow 2.2\sin\frac{B}{2}\cos\frac{B}{2} = 2\sin\frac{A + C}{2}\cos\frac{A - C}{2}$

$\Rightarrow 2\sin\frac{B}{2} = \cos\frac{A - C}{2} \Rightarrow 2\cos\frac{A + C}{2} = \cos\frac{A - C}{2}$

$\cos BIO = \frac{BI^2 + IO^2 - BO^2}{2BI.IO}$

Now $BI^2 + IO^2 - BO^2 = r^2\cosec^2\frac{B}{2} + R^2 - 2rR - R^2$

$= r^2\cosec^2\frac{B}{2} - 2rR$

$= \frac{r.4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{\sin^2\frac{B}{2}} - 2rR$

$= \frac{2rR.2\sin\frac{A}{2}\sin\frac{C}{2}}{\sin\frac{B}{2}} - 2rR$

$= \frac{2rR\left[\cos\frac{A - C}{2} - \cos\frac{A + C}{2}\right]}{\sin\frac{B}{2}} - 2rR$

$= 2rR\frac{2\cos\frac{A + C}{2}- \cos\frac{A + C}{2}}{\sin\frac{B}{2}} - 2rR = 0$

Thus, $\triangle BIO$ is a right angled triangle.

Now the sufficient condition can be proved similarly.
$\frac{A}{2} + \frac{B}{2} = \frac{\pi}{2} - \frac{C}{2}$

$\Rightarrow \cot\left(\frac{A}{2} + \frac{B}{2}\right) = \cot\left(\frac{\pi}{2} - \frac{C}{2}\right)$

$\Rightarrow \frac{\cot\frac{A}{2}\cot\frac{B}{2} - 1}{\cot\frac{A}{2} + \cot\frac{B}{2}} = \tan\frac{C}{2} = \frac{1}{\cot\frac{C}{2}}$

$\Rightarrow \cot\frac{A}{2} + \cot \frac{B}{2} + \cot\frac{C}{2} = \cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}$
The diagram is given below:

$HL = r_2, ID=r \therefore IL = ID - LD = ID - HK = r - r_2$

In $\triangle IHL,$

$\cot\frac{B}{2} = \frac{HL}{IL} = \frac{r_2}{r - r_2}$

Similarly, $\cot\frac{A}{2} = \frac{r_1}{r - r_1}$ and

$\cot\frac{C}{2} = \frac{r_3}{r - r_3}$

Now following the result of previous problem

$\frac{r_1}{r - r_1} + \frac{r_2}{r - r_2} + \frac{r_3}{r - r_3} = \frac{r_1r_2r_3}{(r - r_1)(r - r_2)(r - r_3)}$
The diagram is given below:

Let $O$ be the circumcenter and $P$ the orthocenter of the $\triangle ABC.$ From geometry,

$\angle BOF = \angle COF = A$ and $AP = 2OF = 2R\cos A, BF = R\sin A$

From right angled $\triangle ADB$

$BD = c\cos B, AD = c\sin B$

Now, $PM = AD - (AP + MD) = c\sin B - (2R\cos A + R\cos A) = c\sin B - 3R\cos A$

$OM = FD = BF - BD = R\sin A - c\cos B$

$\therefore \tan\theta = \frac{PM}{OM} = \frac{c\sin B - 3R\cos A}{R\sin A - c\cos B}$

$= \frac{2R\sin C\sin B - 3R\cos A}{R\sin A - 2R\sin C\cos B}$

$= \frac{2\sin B\sin C - 2\cos[\pi - (B + C)]}{\sin[\pi - (B + C)- 2\sin C\cos B]}$

$= \frac{2\sin B\sin C + 3\cos(B + C)}{\sin(B + C) - 2\sin C\cos B}$

$= \frac{3\cos B\cos C - \sin B\sin C}{\cos C\sin B - \sin C\cos B}$

$= \frac{3 - \tan B\tan C}{\tan B - \tan C}$
The diagram is given below:

Let $ABC$ be the triangle. Let $O$ and $I$ be the circumcenter and in-center of the $\triangle ABC.$ Let $P$ be the center and $x,$ the radius of the circle drawn which touches the inscribed and circumscribed circle, of $\triangle ABC$ and the side $BC$ externally. Let us join $OP$ and extend up to $Q.$ Let $ID\perp BC.$ Clearly, $P$ will lie on the extended part of $ID.$ Draw the line $ON$ parallel to the line $IP.$ Join $NP.$

Clearly, $OB = OC OQ = R, OP = OQ - PQ = R - x$

$ON = OM + MN = R\cos A + x$

$NP = MD = OM - CD = \frac{a}{2} - r\cot \frac{C}{2}$

$= \frac{a}{2} - \frac{\Delta}{s}.\frac{s(s - c)}{\Delta} = \frac{a}{2} - (s - c)$

$= \frac{c - b}{2}$

From right angled $\triangle ONP, OP^2 = ON^2 + NP^2$

$(R - x)^2 = (R\cos A + x)^2 + \left(c - b\right)^2$

$R^2 + x^2 - 2Rx = R^2\cos^2A + x^2 + 2Rx\cos A + \left(\frac{c - b}{2}\right)^2$

$2Rx(1 + \cos A) = R^2(1 - \cos^2A) - \left(\frac{c - b}{2}\right)^2$

$4Rx\cos^2\frac{A}{2} = R^2\sin^2A - \left(\frac{b - c}{2}\right)^2 = \frac{a^2}{4} - \frac{(b - c)^2}{4}$

$4Rx\frac{s(s - a)}{bc} = \frac{a +b - c}{2}.\frac{a - b + c}{2} = (s - b)(s - c)$

$x = \frac{\Delta}{a}\tan^2\frac{A}{2}$
The diagram is given below:

Since angles n the same segment of a circle are equal. $\therefore \angle BED= \angle BAD = \frac{A}{2}$

and $\angle BEF = \angle BCF = \frac{C}{2}$

Now $\angle DEF = \angle BEF + \angle BED = \frac{C}{2} + \frac{A}{2} = \frac{C + A}{2} = 90^\circ - \frac{B}{2}$

Similarly, $DFE = 90^\circ - \frac{C}{2}, \angle EDF = 90^\circ - \frac{A}{2}$

Now area of $\triangle DEF = \frac{1}{2}DE.DF.\sin \angle EDF$

Let $R$ be the circum-radius of $\triangle ABC$ then clearly $R$ is also the circum-radius of $\triangle DEF.$

Applying sine rule in $\triangle DEF,$ we have

$\frac{DE}{\sin DFE} = \frac{DF}{\sin DEF} = \frac{EF}{\sin EDF} = 2R$

$\Rightarrow DE = 2R\sin DFE = 2R\sin\left(90^\circ - \frac{C}{2}\right) = 2R\cos\frac{C}{2}$

Similarly, $DF = 2R\cos\frac{B}{2}$

So, area of $\triangle DEF = \frac{1}{2}.4R^2\cos\frac{B}{2}\cos\frac{C}{2}.\sin\left(90^\circ - \frac{A}{2}\right)$

$= 2R^2\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$

$= 2R^2\sqrt{\frac{s(s - a)}{bc}}\sqrt{\frac{s(s - b)}{ca}}\sqrt{\frac{s(s - c)}{ab}}$

$= \frac{2R^2s\sqrt{s(s - a)(s - b)(s - c)}}{abc} = \frac{2R^2s\Delta}{abc} = \frac{R^2.s.4\Delta}{2abc} = \frac{R^2s}{2\frac{abc}{4\Delta}}$

Here, are of $\triangle ABC = \Delta$

$= \frac{R^2s}{2R} = \frac{R}{2}s$

$\Rightarrow \frac{\Delta DEF}{\Delta ABC} = \frac{Rs}{2\Delta} = \frac{R}{2r}$
The diagram is given below:

$\Delta A'B'C' = \Delta ABC - (\Delta AC'B' + \Delta BA'C' + \Delta CA'B')$

$\Delta AC'B' = \frac{1}{2}AC'.AB'\sin A$

$\because CC'$ is the internal bisector of $\angle C$

$\frac{AC'}{C'B} = \frac{AC}{CB} = \frac{b}{a}$

$\Rightarrow AC' = bk, C'B = ak,$ where $k$ is some constant dependent on angles.

$AC' + C'B = AB \Rightarrow c = ak + bk \Rightarrow k = \frac{c}{a + b}$

$\therefore AC' = \frac{bc}{a + b}$

Similarly, $AB' = \frac{bc}{a + c}$

$\Delta AC'B' = \frac{1}{2}\frac{bc}{a + b}\frac{bc}{a + c}\sin A$

Let $\Delta$ be the area of $\triangle ABC,$ then $\Delta = \frac{1}{2}bc\sin A$

$\therefore \Delta AC'B' = \frac{bc}{(a + b)(a + c)}\Delta$

Similalry, $\Delta BA'C' = \frac{ac}{(a + b)(b + c)}\Delta$

and $\Delta CA'B' = \frac{ab}{(a + c)(b + c)}\Delta$

$\therefore \Delta A'B'C' = \Delta.\frac{2abc}{(a + b)(b + c)(c + a)}$

$\therefore \frac{\Delta A'B'C'}{\Delta ABC} = \frac{2\sin A\sin B\sin C}{(\sin A + \sin B)(\sin B + \sin C)(\sin C + \sin A)}$

$= \frac{2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{\cos\frac{A - B}{2}\cos\frac{B - C}{2}\cos\frac{C - A}{2}}$
The diagram is given below:

Let the given triangle be $ABC$ and the similar triangle inscribed in triangle $ABC$ be $A'B'C'$ such that

$A=A', B=B', C=C'$

Let $B'C'=\lambda a, A'C'=\lambda b, A'B' = \lambda c$

According to the quuestion $\angle B'OC=\theta$

Clearly, $\angle OC'B = B - \theta = \angle AC'B'$

$\angle BC'A' = 180^\circ - (B - \theta + C) = A + \theta$

$\angle AB'C' = 180^\circ - (B - \theta + A) = C + \theta$

$\angle A'B'C' = 180^\circ - (C + \theta + B) = A - \theta$

Applying sine rule in the triangle $BC'A',$

$\frac{BA'}{\sin(A +\theta)} = \frac{\lambda b}{\sin B}\Rightarrow BA' = \frac{\lambda b}{\sin B}\sin(A + \theta)$

$\Rightarrow BA' = \lambda 2R\sin(A + \theta)$

Applying sine rule in $A'B'C,$

$\frac{A'C}{\sin(A - \theta)} = \frac{\lambda c}{\sin C}$

$\Rightarrow A'C = \lambda 2R\sin(A - \theta)$

$BC = BA' + A'C$

$\Rightarrow a = \lambda 2R\sin(A + \theta) + \lambda 2R\sin(A - \theta)$

$\Rightarrow a = 2R\lambda[\sin(A + \theta) + \sin(A - \theta)] = 2R\lambda 2\sin A\cos\theta$

$\Rightarrow a = 2\lambda a\cos\theta$

$\Rightarrow 2\lambda\cos\theta = 1$
We have to prove that $r_1 + r_2 + r_3 - r = 4R$

L.H.S. $= \frac{\Delta}{s - a} + \frac{\Delta}{s - b} + \frac{\Delta}{s - c} - \frac{\Delta}{s}$

$= \Delta\left[\frac{s - a + s - b}{(s - a)(s - b)} + \frac{s - s + c}{s(s - c)}\right]$

$=\Delta\left[\frac{c}{(s - a)(s - b)} + \frac{c}{s(s - c)}\right]$

$=\Delta.c \left[\frac{s(s - c) + (s - a)(s - b)}{s(s - a)(s - b)(s - c)}\right]$

$=\Delta.c.\frac{1}{\Delta^2}[s^2 - sc + s^2 - (a + b)s + ab]$

$=\frac{c}{\Delta}[2s^2 - s(a + b + c) + ab] = \frac{c}{\Delta}[2s^2 - 2s^2 + ab]$

$= \frac{abc}{\Delta} = 4R =$ R.H.S.
We have to prove that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$

L.H.S. $= \frac{s - a}{\Delta} + \frac{s - b}{\Delta} + \frac{s - c}{\Delta}$

$= \frac{3s - (a + b + c)}{\Delta} = \frac{3s - 2s}{\Delta} [\because 2s = a + b + c]$

$= \frac{s}{\Delta} = \frac{1}{r} =$ R.H.S.
We have to prove that $\frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r^2} = \frac{a^2 + b^2 + c^2}{\Delta^2}$

L.H.S. $= \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r^2}$

$= \frac{(s - a)^2 + (s - b)^2 + (s - c)^2 + s^2}{\Delta^2}$

$= \frac{4s^2 - 2s(a + b + c) + a^2 + b^2 + c^2}{\Delta^2}$

$=\frac{4s^2 - 2s.2s + a^2 + b^2 + c^2}{\Delta^2}$

$= \frac{a^2 + b^2 + c^2}{\Delta^2} =$ R.H.S.
$r = \frac{\Delta}{s} = \frac{\Delta}{s}\frac{s - a}{s - a}$

We know that $\tan \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}}$ and $\Delta = \sqrt{s(s - a)(s - b)(s - c)}$

$\Rightarrow r = (s - a)\tan\frac{A}{3}$

Similarly, $r = (s - b)\tan\frac{B}{2}, r = (s - c)\tan\frac{C}{2}$
We have to prove that $\frac{1}{\sqrt{A}} = \frac{1}{\sqrt{A_1}} + \frac{1}{\sqrt{A_2}} + \frac{1}{\sqrt{A_3}}$

R.H.S. $= \frac{1}{\sqrt{A_1}} + \frac{1}{\sqrt{A_2}} + \frac{1}{\sqrt{A_3}}$

$= \frac{1}{\sqrt{\pi}}\left(\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}\right)$

$=\frac{1}{\sqrt{\pi}}\left(\frac{s - a}{\Delta} + \frac{s - b}{\Delta} + \frac{s - c}{\Delta}\right)$

$= \frac{1}{\sqrt{\pi}}\left(\frac{3s - (a + b + c)}{\Delta}\right) = \frac{1}{\sqrt{\pi}}\frac{s}{\Delta} = \frac{1}{\sqrt{\pi}}.\frac{1}{r}$

$= \frac{1}{\sqrt{A}} =$ L.H.S.
$\frac{r_1}{bc} + \frac{r_2}{ca} + \frac{r_3}{ab} = \frac{s\tan\frac{A}{2}}{bc} + \frac{s\tan\frac{B}{2}}{ca} = \frac{s\tan\frac{C}{2}}{ab}$

$= \frac{s}{abc}\left(a\tan\frac{A}{2} + b\tan\frac{B}{2} + c\tan\frac{C}{2}\right)$

$= \frac{s}{abc}\left(2R\sin A\tan\frac{A}{2} + 2R\sin B\tan\frac{B}{2} + 2R\sin C\tan\frac{C}{2}\right)$

$= \frac{s}{abc}.4R\left(\sin^2\frac{A}{2} + \sin^2\frac{B}{2} + \sin^2\frac{C}{2}\right)$

$= \frac{s}{\Delta}\left(\frac{1 - \cos A}{2} + \frac{1 - \cos B}{2} + \frac{1 - \cos C}{2}\right)$

$= \frac{1}{r}\left(\frac{3}{2} - \frac{\cos A + \cos B + \cos C}{2}\right)$

We know that $\cos A + \cos B + \cos C = 1 + \frac{r}{R}$

$= \frac{1}{r}\left(1 - \frac{r}{2R}\right) = \frac{1}{r} - \frac{1}{2R} =$ R.H.S.
Let $D$ be the point where perpendicular from $A$ meets $BC.$ Then $AD = h$

The diagram is given below:

Clearly, $OB = r, AD = h, OD=h - r$ (If $O$ is below $BD$ then $OD = r - h$ )

$BD = \sqrt{OB^2 - OB^2} = \sqrt{r^2 - (h - r)^2} = \sqrt{2rh - h^2}$

Area of triangle $= \frac{1}{2}.2.BD.h = h\sqrt{2rh - h^2}$
Let the sides be $a, b, c$ then $\Delta = \frac{1}{2}ap_1 = \frac{1}{2}bp_2 = \frac{1}{2}cp_3$

$\Rightarrow p_1 = \frac{2\Delta}{a}, p_2 = \frac{2\Delta}{b}, p_3 = \frac{2\Delta}{c}$

L.H.S. $= \frac{\cos A}{p_1} + \frac{\cos B}{p_2} + \frac{\cos C}{p_3}$

$= \frac{1}{2\Delta}[a\cos A + b\cos B + c\cos C]$

$= \frac{2R}{2\Delta}[\sin A\cos A + \sin B\cos B + \sin C\cos C]$

$= \frac{R}{2\Delta}[\sin 2A + \sin 2B + \sin 2C] = \frac{R}{2\Delta}4\sin A\sin B\sin C$

$= \frac{abc}{4\Delta}.\frac{1}{R^2} = \frac{R}{R^2} = \frac{1}{R} =$ R.H.S.
This has been already proved in 149.
L.H.S. $= r_1r_2r_3 = \frac{\Delta}{s - a}\frac{\Delta}{s - b}.\frac{\Delta}{s - c}$

$= \frac{\Delta^3}{(s - a)(s - b))(s - c)} = \Delta .s$

R.H.S. $= r^3\cot^2\frac{A}{2}\cot^2\frac{B}{2}\cot^2\frac{C}{2}$

$= \frac{\Delta^3}{s^3}.\frac{s^2(s - a)^2}{\Delta^2}.\frac{s^2(s - b)^2}{\Delta^2}.\frac{s^2(s - c)^2}{\Delta^2}$

$= \frac{s^3(s - a)^2(s - b)^2(s - c)^2}{\Delta^3} = \Delta .s$
We have to prove that $a(rr_1 + r2r_3) = b(rr_2 + r_3r_1) = c(rr_3 + r_1r_2) = abc$

$a(rr_1 + r_2r_3) = a\left(\frac{\Delta}{s}.\frac{\Delta}{s - a} + \frac{\Delta}{(s - b)}\frac{\Delta}{s - c}\right)$

$= \Delta^2.a\left(\frac{(s - b)(s - c) + s(s - a)}{s(s - a)(s - b)(s - c)}\right)$

$= \Delta^2.a\left(\frac{2s^2 - s(a + b + c) + bc}{\Delta^2}\right)$

$= abc$

Similalry other terms can be evaluated to same value of $abc.$
We have to prove that $(r_1 + r_2)\tan\frac{C}{2} = (r_3 - r)\cot\frac{C}{2} = c$

$(r_1 + r_2)\tan\frac{C}{2} = \left(\frac{\Delta}{s - a} + \frac{\Delta}{s - b}\right)\frac{(s - a)(s - b)}{\Delta}$

$= s - b + s - a = c$

$(r_3 - r)\cot\frac{C}{2} = \left(\frac{\Delta}{s - c} - \frac{\Delta}{s}\right)\frac{\Delta}{(s - a)( s - b)}$

$= \Delta^2\left(\frac{s - s + c}{s(s - a)(s - b)(s - c)}\right) = c$
We have to prove that $4R\sin A\sin B\sin C = a\cos A + b\cos B + c\cos C$

R.H.S. $= R(2\sin A\cos A + 2\sin B\cos B + 2\sin C\cos C) = R(\sin 2A + \sin 2B + \sin 2C)$

$= R(2\sin(A + B)\cos(A - B) + 2\sin C\cos C) = 2R(\sin C\cos(A - B) + \sin C\cos C)[\because \sin(A + B) = \sin(\pi - C) = \sin C]$

$= 2R\sin C[\cos(A - B) - \cos(A + B)][\because \cos C = \cos(\pi - A - B) = -\cos(A + B)]$

$= 2R\sin C. 2\sin A\sin B = 4R\sin A\sin B\sin C =$ L.H.S.
We have to prove that $(r_1 - r)(r_2 - r)(r_3 - r) = 4Rr^2$

L.H.S. $= \left(\frac{\Delta}{s - a} - \frac{\Delta}{s}\right)\left(\frac{\Delta}{s - b} - \frac{\Delta}{S}\right)\left(\frac{\Delta}{s - c} - \frac{\Delta}{s}\right)$

$= \Delta^3\left(\frac{s - s + a}{s(s - a)}\right)\left(\frac{s - s + b}{s(s - b)}\right)\left(\frac{s - s + c}{s(s - c)}\right)$

$= \Delta^3.\frac{abc}{s^3(s - a)(s - b)(s - c)} = \frac{\Delta^3.abc}{s^2\Delta^2} = \frac{abc.\Delta}{s^2}$

$= \frac{abc}{\Delta}.\frac{\Delta^2}{s^2} = 4Rr^2 =$ R.H.S.
We have to prove that $r^2 + r_1^2 + r_2^2 + r_3^2 = 16R^2 - a^2 - b^2 - c^2$

$(r_1 + r_2 + r_3 - r)^2 = r_1^2 + r_2^2 + r_3^2 + r^2 - 2r(r_1 + r_2 + r_3) + 2(r_1r_2 + r_2r_3 + r_3r_1)$

We know that $r_1 + r_2 + r_3 - r = 4R$ and $(r_1r_2 + r_2r_3 + r_3r_1) = s^2$

$r(r_1 + r_2 + r_3) = \frac{\Delta}{s}\left(\frac{\Delta}{s -a} + \frac{\Delta}{s - b} + \frac{\Delta}{s - c}\right)$

$= \frac{\Delta^2}{s(s - a)} + \frac{\Delta^2}{s(s - b)} + \frac{\Delta^2}{s(s - c)} = -s^2 + (ab + bc + ca)$

$16R^2 = r_1^2 + r_2^2 + r_3^2 + r^2 -2[-s^2 + (ab + bc + ca)] + 2s^2$

$\Rightarrow r^2 + r_1^2 + r_2^2 + r_3^2 = 16R^2 - a^2 - b^2 - c^2$
We know that $IA = r\cosec\frac{A}{2}, IB=r\cosec\frac{B}{2}, IC=r\cosec\frac{C}{2}$

L.H.S. $= \frac{r^3}{\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}} = \frac{r^3.4R}{4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}$

$= \frac{r^3.4R}{r} = 4R.r^2 = 4R.\frac{\Delta^2}{s^2} = \frac{abc\Delta}{s^2}$

R.H.S. $= abc.\frac{(s - a)^2(s - b)^2(s - c)^2}{\Delta^3} = \frac{abc\Delta}{s^2}$
From here we can say that $AI_1 = r_1\cosec\frac{A}{2}$
$II_1 = AI_1 - AI = r_1\cosec\frac{A}{2} - r\cosec\frac{A}{2}$

$= \left(\frac{\Delta}{s - a} - \frac{\Delta}{s}\right)\cosec\frac{A}{2}$

$= \Delta .\frac{a}{s(s - a)}\sqrt{\frac{bc}{(s - b)(s - c)}} = a\sec\frac{A}{3}$
If $E_2$ be the point of contact of the circle whose center is $I_2$ with the side $AC$ of the triangle $ABC,$ we have

$AI_2 = AE_2\sec I_2AE_2 = AE_2sec\left(90^\circ - \frac{A}{2}\right) = (s - b)\cosec \frac{A}{2}$

$I_2I_3 = AI_2 + AI_3 = (s - b + s - c)\cosec\frac{A}{2} = a\cosec\frac{A}{2}$
We have deduced that $II_1 = a\sec\frac{A}{2}$ in problem 176. So $II_2 = b\sec\frac{B}{2}$ and $II_3 = c\sec\frac{C}{2}$

L.H.S. $= II_1.II_2.II_3 = abc\sec\frac{A}{2}\sec\frac{B}{2}\sec\frac{C}{2}$

$= 8R^3\frac{2\sin\frac{A}{2}\cos\frac{A}{2}.2\sin\frac{B}{2}\cos\frac{B}{2}.2\sin\frac{C}{2}\cos\frac{C}{2}}{\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}}$

$= 16R^2.4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} = 16R^2r =$ R.H.S.
$II_1 = a\sec\frac{A}{2}, I_2I_3 = a\cosec\frac{A}{2}$

$II_1^2 + I_2I_3^2 = a^2\left(\frac{1}{\sin^2\frac{A}{2}} + \frac{1}{\sin^2\frac{A}{2}}\right)$

$= \left(\frac{a}{\sin\frac{A}{2}\cos\frac{A}{2}}\right)^2 = \left(\frac{2a}{2\sin\frac{A}{2}\cos\frac{A}{2}}\right)^2$

$= 16R^2 [\because a = 2R\sin A]$

Similalrly other terms can be proven to be equal to $16R^2$
We know that $OI^2 = R^2\left(1 - 8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right)$

$= R^2\left[1 - 4\left(\cos \frac{A - B}{2} - \cos\frac{A + B}{2}\right)\sin\frac{C}{2}\right]$

$= R^2\left[1 - 4\cos\frac{A - B}{2}\cos\frac{A + B}{2} + 4\sin^2\frac{C}{2}\right][\because \sin \frac{C}{2} = \cos\frac{A + B}{2}]$

$= R^2\left[1 - 2(\cos A + \cos B) + 2(1 - \cos C)\right]$

$=R^2(3 - 2\cos A - 2\cos B - 2\cos C)$
We have, $IH^2 = AH^2 + AI^2 - 2.AH.AI.\cos IAH$

$\angle IAH = \frac{A}{2} - \angle HAC = \frac{A}{2} - (90^\circ - C) = \frac{C - B}{2}$

$IH^2 = 4R^2\cos^2A + 16R^2\sin^2\frac{B}{2}\sin^2\frac{C}{2} - 16R^2\cos A\sin\frac{B}{2}\sin\frac{C}{2}\cos\frac{C - B}{2}$

$= 4R^2\left[\cos^2A + 4\sin^2\frac{B}{2}\sin^2\frac{C}{2} - 4\cos A\sin\frac{B}{2}\sin\frac{C}{2}\cos\frac{C}{2}\cos\frac{B}{2} - 4\cos A\sin^2\frac{B}{2}\sin^2\frac{C}{2}\right]$

$= 4R^2\left[\cos^2A + 4\sin^2\frac{B}{2}\sin^2\frac{C}{2}(1 - \cos A) - \cos A\sin B\sin C\right]$

$= 4R^2\left[8\sin^2\frac{A}{2}\sin^2\frac{B}{2}\sin^2\frac{C}{2} + \cos^2A - \cos A\sin B\sin C\right]$

$= 2r^2 + 4R^2\cos A(\cos A - \sin B\sin C)$

$= 2r^2 - 4R^2\cos A\cos B\cos C$
We know that $OG = \frac{1}{3}OH \Rightarrow OG^2 = \frac{OH^2}{9}$

$= \frac{1}{9}[R^2 - 8R^2\cos A\cos B\cos C] = \frac{R^2}{9}[1 - 4\{\cos(A + B) + \cos(A - B)\}\cos C]$

$= \frac{R^2}{9}[1 + 4\cos^2C + 4\cos(A - B)\cos(A + B)]$

$= \frac{R^2}{9}[1 + 2(1 + \cos 2C) +2(\cos 2A + \cos 2C)]$

$= \frac{R^2}{9}[3 + \cos 2A + \cos 2B + \cos 2C]$

$= \frac{R^2}{9}[9 - 2(1 - \cos 2A) - 2(1 - \cos 2B) - 2(1 - \cos 2C)]$

$= \frac{R^2}{9}[9 - 4(\sin^2A + \sin^2B + \sin^2C)]$

$= R^2 - \frac{1}{9}(2R\sin A)^2 - \frac{1}{9}(2R\sin B)^2 - \frac{1}{9}(2R\sin C)^2$

$= R^2 - \frac{1}{9}(a^2 + b^2 + c^2)$
The diagram is given below:

Clearly, $R = \frac{b}{2\sin\frac{\alpha}{2}} = \frac{b\cosec\frac{\alpha}{2}}{2}$
We know that in a $\triangle ABC,$ $r = 4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$

Let $AD$ be the perpedicular bisector to $BC.$

$\Delta = BD.AD = b\cos\alpha.b\sin\alpha = \frac{1}{2}b^2\sin2\alpha$

$r = \frac{\Delta}{s} = \frac{\frac{1}{2}b^2\sin2\alpha}{\frac{1}{2}(b + b + 2b\cos\alpha)} = \frac{b\sin2\alpha}{2(1 + \cos\alpha)}$
$OI = |OD + DI| = |OD + r|$ because $\alpha < \pi/4, A>\pi/2$ and $O$ lies on $AD$ produced.

From right-angled $\triangle ODB,$ we get

$OD^2 = OB^2 - BD^2 = R^2 - b^2\cos^2\alpha$

$= \frac{1}{4}\frac{b^2}{\sin^2\alpha} - b^2\cos^2\alpha$

$= \frac{b^(1 - 4\sin^2\alpha\cos^2\alpha)}{4\sin^2\alpha} = \frac{b^2(\cos^2\alpha - \sin^2\alpha)}{4\sin^2\alpha}$

$= \frac{b^2\cos^22\alpha}{(2\sin\alpha)^2}$

$\therefore OI = \left|\frac{b\sin2\alpha}{2(1 + \cos\alpha)} + \frac{b\cos2\alpha}{2\sin\alpha}\right|$

$= \left|\frac{b\sin2\alpha}{4\cos^2\frac{\alpha}{2}} + \frac{b\cos2\alpha}{4\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}\right|$

$= \left|\frac{b}{4\cos\alpha/2}.\frac{\sin2\alpha\sin\alpha/2 + \cos2\alpha\cos\alpha/2}{\sin\alpha/2\cos\alpha/2}\right|$

$= \left|\frac{b\cos3\alpha/2}{2\sin\alpha\cos\alpha/2}\right|$
L.H.S. $= \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{a + b + c}{abc}$

$= \frac{2s}{4RS} = \frac{1}{2RS/s} = \frac{1}{2Rr} =$ R.H.S.
We know that $r_1 = \frac{\Delta}{s - a}, r_2 = \frac{\Delta}{s - b}, r_3 = \frac{\Delta}{s - c}$ and $r = \frac{\Delta}{s}$

L.H.S. $= \frac{3\Delta}{(s - a)(s - b)(s - c)} = \frac{3s\Delta}{\Delta^2} = \frac{3s}{\Delta} = \frac{3}{r} =$ R.H.S.
The diagram is given below:

$2s = 2(\alpha + \beta + \gamma) \Rightarrow s = \alpha + \beta + \gamma$

We know that $r = \frac{\Delta}{s} \Rightarrow s^2 = \frac{(s - a)(s - b)(s - c)}{s}$

Clearly, $s - a = \gamma, s - b = \beta, s - c = \alpha$

$\Rightarrow s = \frac{\alpha\beta\gamma}{\alpha + \beta + \gamma}$
The diagram is given below:

Let $PQ=x, PQ\parallel BC, RS = y, RS\parallel AC, TU = z, TU\parallel AB$

In $\triangle APQ, \frac{x}{\sin A} = \frac{AQ}{\sin B} = \frac{AP}{\sin C}$

$\Rightarrow AQ = \frac{bx}{a}, AP = \frac{cx}{a}$

$r = \left(\frac{x + AP + AQ}{2}\right)\tan\frac{A}{2} = \frac{a + b + c}{2}x\tan\frac{A}{2}$

$= \frac{sx}{a}\tan\frac{A}{2} = (s - a)\tan\frac{A}{2}$

$\Rightarrow \frac{sx}{a} = s - a$

Similarly, $\frac{sy}{b} = s - b$ and $\frac{sz}{c} = s - c$

$\Rightarrow s\left(\frac{x}{a} + \frac{y}{b} + \frac{z}{c}\right) = 3s - (a + b + c)$

$\Rightarrow \frac{x}{a} + \frac{x}{b} + \frac{x}{c} = 1$
The diagram is given below:

Since $I$ is the incenter, $AI$ will be angle bisector. Let $AI$ cut circumcirlce at $D.$

$\angle DBI=\angle DBC+\angle IBC=\angle DAB+\angle ABI=\angle BID$ and then $DB=DI$

Likewise $DC = DI$ and then $DB = BI = DC$

$I_1C$ bisects $\angle BCT \Longrightarrow \angle ICI_1 = 90^\circ$

Let the perpendicular bisector of $BC$ cut circumcircle at $M$ also.

$\triangle SAI_1 ~ \triangle BMD$

Power of $I_1$ w.r.t the circumcircle of $\triangle ABC = OI_1^2 - R^2 = I_1D.I_1A$

$\Rightarrow \frac{MD}{BD} = \frac{I_1A}{SI_1} \Rightarrow 2Rr_1 = OI_1^2 - R^2$

Thus, $OI_1 = R^2 + 2Rr_1$

Thus length of tangent $t_1^2 = OI_1^2 - R^2 = 2Rr_1$

$\frac{1}{t_12} + \frac{1}{t_2^2} + \frac{1}{t_3^2} = \frac{1}{2R}\left(\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}\right)$

$= \frac{1}{2R}\frac{3s - (a + b + c)}{\Delta} = \frac{4\Delta}{2}.\frac{s}{\Delta} = \frac{2s}{abc}$
$r_1 = \frac{\Delta}{s - a}$ and so on. Given,

$\left(1 - \frac{r_1}{r_2}\right)\left(1 - \frac{r_1}{r_3}\right) = 2$

$\left(1 - \frac{s - b}{s - a}\right)\left(1 - \frac{s - c}{s - a}\right) = 2$

$(b - a)(c - a) = 2(s - a)^2$

$bc + a^2 - ac - ab = (b + c - a)^2/2$

$b^2 + c^2 = a^2$

Thus the triangle is right-angled.
$\frac{\text{Area of in-circle}}{\text{Area of triangle}} = \frac{\pi r^2}{\Delta} = \frac{\pi}{\Delta}.\frac{\Delta^2}{s^2}$

$= \pi .\frac{\Delta}{s^2}$

$\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2} = \sqrt{\frac{s(s - a)}{(s - b)(s - c)}.\frac{s(s - b)}{(s - a)(s - c)}.\frac{s(s - c)}{(s - a)(s - b)}}$

$= \sqrt{\frac{s^4}{s(s - a)(s - b)(s - c)}} = \frac{s^2}{\Delta}$

$\Rightarrow \frac{\pi}{\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}} = \frac{\pi\Delta}{s^2}$

Thus, we have the desired result by combining both the equations.
The diagram is given below:

Let $O$ be the center of the regular polygon $A_1, A_2, A_3, \ldots, A_n$ which has $n$ sides.

Since it is a regular polyogn so $\angle A_1OA_2 = \angle A_2OA_3 = \ldots = \angle A_nOA_1 = \frac{2\pi}{n}$

Also, let $OA_1 = OA_2 = \ldots = OA_n = r$

Applying cosine rule in $\triangle A_1OA_2,$

$\cos\frac{2\pi}{n} = \frac{OA_1^2 + OA_2^2 - A_1A_2^2}{2OA_1.OA_2}$

$\Rightarrow A_1A_2^2 = 2r^2\left(1 - \cos \frac{2\pi}{n}\right)$

$\Rightarrow A_1A_2^2 = 4r^2\sin^\frac{\pi}{n}$

$\Rightarrow A_1A_2 = 2r\sin\frac{\pi}{n}$

Likewise $A_1A_3 = 2r\sin\frac{2\pi}{n} A_1A_4 = 2r\sin\frac{3\pi}{n}$

Given, $\frac{1}{A_1A_2} = \frac{1}{A_1A_3} + \frac{1}{A_1A_4}$

$\Rightarrow \frac{1}{2r\sin\frac{\pi}{n}} = \frac{1}{2r\sin\frac{2\pi}{n}} + \frac{1}{2r\sin\frac{3\pi}{n}}$

$\Rightarrow \frac{1}{\sin\frac{\pi}{n}} - \frac{1}{\sin\frac{3\pi}{n}} = \frac{1}{\sin\frac{2\pi}{n}}$

$\Rightarrow \frac{2\cos\frac{2\pi}{n}\sin\frac{\pi}{n}}{\sin\frac{\pi}{n}\sin\frac{3\pi}{n}} = \frac{1}{\sin\frac{2\pi}{n}}$

$\Rightarrow 2\cos\frac{2\pi}{n}\sin\frac{2\pi}{n} = \sin\frac{3\pi}{n}$

$\Rightarrow \sin\frac{4\pi}{n} = \sin\frac{3\pi}{n}$

$\Rightarrow \cos\frac{7\pi}{2n}.\sin\frac{\pi}{2n} = 0$

$\Rightarrow \cos\frac{7\pi}{2n} = 0\left[\sin\frac{\pi}{2n} \neq 0\right]$

$\Rightarrow \frac{7\pi}{2n} = \text{odd integer} \times \frac{\pi}{2}$

$\Rightarrow n = \frac{7}{\text{odd integer}} = 7[n\in I, n > 1]$
The diagram is given below:

Let $O$ be the center of the circumscribing circle regular polygon $A_1, A_2, A_3, \ldots, A_n$ which has $n$ sides.

Since the polygon is regular, therefore $O$ will also be the center of inscribing circle.

Let $OD\perp A_1A_2.$ Now $\angle A_1OA_2 = \frac{2\pi}{n}$

$\angle A_OD = \angle A_2OD = \frac{\pi}{n}$

Also, $A_1D = A_2D = a/2$ where $a$ is the length of a side of the polygon.

Here, $R =$ radius of the circumscribing circle $= OA$

and $r =$ radius of the inscribing circle $= OD$

From right angled triangle $ODA_1$

$\sin\frac{\pi}{n} = \frac{a/2}{R} \Rightarrow R = \frac{a}{2}\cosec\frac{\pi}{n}$

and $\tan\frac{\pi}{n} = \frac{a/2}{r} \Rightarrow r = \frac{a}{2}\cot\frac{\pi}{n}$

$\therefore R + r = \frac{a}{2}\left[\cosec\frac{\pi}{n} + \cot\frac{\pi}{n}\right]$

$= \frac{a}{2}\left[\frac{1 + \cos\frac{\pi}{n}}{\sin\frac{\pi}{n}}\right]$

$= \frac{a}{2}\cot\frac{\pi}{2n}$
Let $ABCD$ be a quadrilateral such that $AB = 3 cm, BC = 4 cm, CD= 5 cm$ and $AD = 6 cm.$

Also, let that $\angle BAC = \theta$ and $\angle BCD = 120^\circ - \theta$ as it is given that sum of pair of opposite angles is $120^\circ.$

Applying cosine law in $\triangle ABD,$

$\cos\theta = \frac{AB^2 + AD^2 - BD^2}{2.AB.AD} \Rightarrow BD = 45 - 36\cos\theta$

Applying cosine law in $\triangle BCD,$

$\cos(120^\circ - \theta) = \frac{BC^2 + CD^2 - BD^2}{2.BC.CD} \Rightarrow BD^2 = 41 + 20\cos \theta - 20\sqrt{3}\sin\theta$

Thus, $45 - 36\cos\theta = 41 + 20\cos\theta - 20\sqrt{3}\sin\theta$

$\Rightarrow 14\cos\theta - 5\sqrt{3}\sin\theta = 1$

Area of the quadrilateral $= \Delta ABD + \Delta BCD$

$= \frac{1}{2}3.6.\sin\theta + \frac{1}{2}4..5\sin(120^\circ - \theta)$

$= 14\sin\theta + 5\sqrt{3}\sin\theta = z$ (let)

Solving the two equations thus obtaiined, we get

$196(\sin^2\theta + \cos^2\theta) + 75(\cos^2\theta + \sin^2\theta) = z^2 + 1$

$\Rightarrow z = 2\sqrt{30}$ sq.cm.
Let $ABCD$ be a cyclic quadrilateral such that $AD = 2, AB = 5, \angle DAB = 60^\circ$

Since the quadrilateral is cyclic $\angle BCD = 120^\circ$

Area of quadrilateral $ABD = \frac{1}{2}.2.5.\sin60^\circ = \frac{5\sqrt{3}}{2}$

Area of $\triangle BCD =$ Area of quadrilateral $ABCD$ - Area of $\triangle ABD$

$= 4\sqrt{3} = \frac{5\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$

Let $CD = x, BC = y$

Now area of $\triangle BCD = \frac{1}{2}.x.y.\sin120^\circ \Rightarrow \frac{3\sqrt{3}}{2} = \frac{1}{2}xy\frac{\sqrt{3}}{2}$

$\Rightarrow xy = 6$

Applying cosine rule in $\triangle ABD,$

$\cos60^\circ = \frac{AD^2 + AB^2 - BD^2}{2.AD.AB} \Rightarrow BD^2 = 19$

Applying cosine rule in $\triangle BCD,$

$\cos120^\circ = \frac{x^2 + y^2 - 19}{2xy}\Rightarrow x^2 + y^2 = 13$

$(x + y)^2 = 25 \Rightarrow x + y = \pm5$

$(x - y)^2 = 1 \Rightarrow x - y = \pm1$

$\Rightarrow x = 3, y = 2$ or $x = 2, y = 3$
From question, $AB = 1, BD = \sqrt{3}.$ Let $BAD=\theta, AD = x, BC = y$ and $CD = z.$ Since the given circle is also circum-circle of $\triangle ABD, \Rightarrow \frac{BD}{\sin\theta} = 2R$

$\Rightarrow \sin\theta = \frac{\sqrt{3}}{2}\Rightarrow \theta = 60^\circ$

Now $\angle BCD = 180^\circ - 60^\circ = 120^\circ$

Applying cosine rule in $\triangle ABD,$

$\cos60^\circ = \frac{AB^2 + AD^2 - BD^2}{2.AB.AD} \Rightarrow \frac{1}{2} = \frac{1 + x^2 - 3}{2x}$

$\Rightarrow x^2 - x - 2 = 0 \Rightarrow x = 2$

Applying cosine law in $\triangle BCD,$

$\cos120^\circ = \frac{y^2 + z^2 - 3}{2yz} \Rightarrow y^2 + z^2 + yz = 3$

Area of quadrilateral $ABCD =$ Area of $\triangle BCD$ + Area of $\triangle ABD$

$\frac{3\sqrt{3}}{2} = \frac{1}{2}1.x.\sin60^\circ + \frac{1}{2}yz\sin 120^\circ$

$\Rightarrow yz = 1$

$\Rightarrow y^2 + z^2 = 2$

$\Rightarrow (y + z)^2 = 4, (y - z)^2 = 0$

$\Rightarrow y = z = 1$
Let $ABCD$ be the cyclic quadrilateral in which $AB = a, BC = b, CD = c, DA =d.$

Applying cosine rule in $\triangle ABC,$

$\cos B = \frac{a^2 + b^2 - AC^2}{2ab} \Rightarrow AC^2 = a^2 + b^2 - 2ab\cos B$

Applying cosine rule in $\triangle ADC,$

$\cos(\pi - B) = \frac{c^2 + d^2 - AC^2}{2cd} \Rightarrow AC^2 = c^2 + d^2 + 2cd\cos B$

$\Rightarrow \cos B = \frac{a^2 + b^2 - c^2 - d^2}{2(ab + cd)}$

$\tan^2\frac{B}{2}= \frac{1 - \cos B}{1 + \cos B}$

$\Rightarrow \tan\frac{B}{2} = \sqrt{\frac{(S - a)(S - b)}{(S - c)(S - d)}}$
Let $ABCD$ be the quadrilateral for which $Ab = a, BC=b, CD=c, DA = d.$ Let diagonals be, $AC=x, BD = y.$

Given, $\angle AOD = \angle BOC = \alpha \therefore \angle AOB = \angle COD = 180^\circ - \alpha$

Area of the quadrilateral $= \frac{1}{2}xy\sin\alpha$

Applying cosine rule in $\triangle AOB,$

$a^2 = AO^2 + BO^2 - 2.AO.BO.\cos(180^\circ - \alpha)$

$a^2 = AO^2 + BO^2 + 2.AO.BO.\cos\alpha$

Likewise in $\triangle BOC,$

$b^2 = BO^2 + CO^2 + 2.BO.CO.\cos \alpha$

And in $\triangle COD$

$c^2 = CO^2 + DO^2 + 2.CO.DO.\cos\alpha$

And in $\triangle AOD,$

$d^2 = AO^2 + DO^2 + 2AO.DO.\cos\alpha$

$a^2 + c ^2 - b^2 - d^2 = 2\cos\alpha.x.y$

Thus, area $= \frac{1}{2}(a^2 + c^2 - b^2 - d^2)$
If the quadrilateral $ABCD$ can have a circle inscribed such that it touches the quadrilateral on sides $AB, BC, CD, DA$ at points $P,Q,R,S$ then we will have

$AP = AS, BP = BQ, CQ = CR, DR = DS$

Since lengths of tnagents are equal,

$\therefore AP + BP + CR + DR = AS + BQ + CQ + DS$

$\Rightarrow AB + CD = AD + BC$

$\Rightarrow a + c = b + d$

$\Rightarrow s = a + c = b + d$

Area of cyclic quadrilateral $= \sqrt{(s - a)(s - b)(s - c)(s - d)} = \sqrt{abcd} = \frac{1}{2}r(a + b + c + d)$ where $r$ is the in-radius.

$r = \frac{2\sqrt{abcd}}{a + b + c + d}$