Subject: Mathematics — Circles
Question: Tangents are parallel to the three sides are drawn to
the in-circle. If $x,y,z$ are the lengths of the parts of the tangents
within the triangle then find the value of
$$\frac{x}{a} + \frac{y}{b} + \frac{z}{c}$$
Answer: 1
Solution: Let
$$PQ = x,\quad PQ \parallel BC,$$
$$SR = y,\quad SR \parallel AC,$$
$$TU = z,\quad TU \parallel AB.$$
Let $I$ be the in-centre of $\bigtriangleup ABC$.
[IMAGE] Consider $\bigtriangleup APQ$.
By sine rule,
$$\frac{x}{\sin A} = \frac{AQ}{\sin B} = \frac{AP}{\sin C}.$$
Therefore,
$$AQ = \frac{\sin B}{\sin A}x = \frac{b}{a}x,$$
$$AP = \frac{\sin C}{\sin A}x = \frac{c}{a}x.$$
Also, $r$ will be ex-radius of $\bigtriangleup APQ$.
$$r = \left( \frac{x + PA + AQ}{2} \right)\tan\frac{A}{2} = \left\lbrack \frac{x + \frac{bx}{a} + \frac{cx}{a}}{2} \right\rbrack\tan\frac{A}{2}.$$
$$r = \frac{(a + b + c)}{2a}x\tan\frac{A}{2} = \frac{sx}{a}\tan\frac{A}{2}.$$
Similarly, in $\bigtriangleup BRS$,
$$BR = \frac{c}{b}y,\quad BS = \frac{a}{b}y\quad\text{and}\quad r = \frac{sy}{b}\tan\frac{B}{2}.$$
$$\left\{ \because\ BR + BS + y = \frac{c}{b}y + \frac{a}{b}y + y = \frac{sy}{b} \right\}.$$
And in $\bigtriangleup CTU$,
$$CT = \frac{a}{c}z,\quad CU = \frac{b}{c}z\quad\text{and}\quad r = \frac{sz}{c}\tan\frac{C}{2}.$$
We also know that,
$$r = (s - a)\tan\frac{A}{2} = (s - b)\tan\frac{B}{2} = (s - c)\tan\frac{C}{2}.$$
Therefore,
$$\frac{sx}{a}\tan\frac{A}{2} = (s - a)\tan\frac{A}{2} \Rightarrow \frac{sx}{a} = s - a,$$
$$\frac{sy}{b}\tan\frac{B}{2} = (s - b)\tan\frac{B}{2} \Rightarrow \frac{sy}{b} = s - b,$$
$$\frac{sz}{c}\tan\frac{C}{2} = (s - c)\tan\frac{C}{2} \Rightarrow \frac{sz}{c} = s - c.$$
On adding,
$$s\left( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} \right) = 3s - (a + b + c) = s.$$
Therefore,
$$\left( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} \right) = 1.$$