prove that sum of perimeters of APQ,BRS,CTV is equal to perimeter of ABC

Question

Question -

ABC is a triangle.PQ,RS,TV are tangents to incircle parallel to BC,CA,AB respectively. prove that sum of perimeters of APQ,BRS,CTV is equal to perimeter of ABC ...

My try -

So first I cancel common sides and wanted to prove that PQ+RS+TV=PR+ST+VQ

But not able to use the given condition that tangents are parallel to corresponding sides...

Any help will be greatly helpful.. Thanks

Source : CTPCM Olympiad book

Answer 1

Let $X, Y, Z$ be the points where the incircle touches $BC, AC, AB$ respectively, and $K, L, M$ the points where the incircle touches the tangents $PQ, RS, TV$, respectively. Now, note that: $PK=PZ, QK=QY, RZ=RL, SX=SL, TX=TM, VM=VY$, so:

$$\begin{array}{rcl}PQ+RS+TV&=&(PK+QK)+(RL+SL)+(VM+TM)\\&=&PZ+QY+RZ+SX+VY+TX\text{ (as per above equalities)}\\&=&(PZ+RZ)+(SX+TX)+(QY+VY)\\&=&PR+ST+VQ\end{array}$$

Answer 2

Let $p$ and $h_a$ be the perimeter and the height of the triangle ABC, and $p_a$ and $h_a'$ be the perimeter and the height of the triangle APQ, respectively. Note the triangles APQ and ABC are similar. Then, with $r$ the incircle radius,

$$\frac{p_a}p = \frac{h_a'}{h_a}=1-\frac{2r}{h_a}\tag 1$$

Also note that

$$Area_{ABC} = \frac12rp = \frac12ah\implies \frac r{h_a} = \frac {a}p$$

where $a = BC$. Substitute into (1) to get

$$p_a = p- 2a$$ Similarly,

$$p_b = p- 2b, \>\>\>\>\>\>\>p_c = p- 2c$$

Thus,

$$p_a + p_b +p_c = 3p- 2(a+b+c) = p$$