Two circles inscribed in isosceles triangle

Question

$$\triangle ABC :AC =BC$$ $$P \in AB$$

$k(O;r)$ is inscribed in $\triangle APC$; $k_2(O_2; r_2)$ is inscribed in $\triangle BPC$;

$D, G$ are points of contact of the circles with $CP$.

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Show that $DG=\frac {|AP - BP|}{2}$.

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I think that it will be helpful if we find point $A_1$, such that $BA_1 = AP$. Now we have to prove $2DG=PA_1$.

Answer 1

Hint:

Use the following fact/theorem.

If $ZX$ and $ZY$ are tangents from $Z$ to a circle so that $X,Y$ are touching points, then $ZX = ZY$.

Now it sholud be easy.

Another hint: Try to prove $$PG= PK = {PB+PC-BC\over 2}$$

Answer 2

Let $|AC|=|BC|=a$, $|AB|=c$, $|AP|=m$, $|CP|=d$.

\begin{align} |CD|&=|CE|=a-|AE| ,\quad |CG|=|CF|=a-|BF| ,\\ |DG|&=\Big||CD|-|CG|\Big| = \Big|a-r\cot\tfrac\alpha2-(a-r_1\cot\tfrac\alpha2)\Big| = \left|r\cot\tfrac\alpha2-r_1\cot\tfrac\alpha2\right| \\ &= \Big|r\cdot\frac{a+m-d}{2r}-r_1\cdot\frac{a+c-m-d}{2r_1}\Big| =\Big|m-\tfrac12 c\Big| =\tfrac12\Big||AP|- |BP|\Big| . \end{align}

Answer 3

(1) Theorem:- Let A’B’C’ be any triangle with corresponding sides a’, b’, c’ and its semi-perimeter = s’. If its in-circle touches A’B’ at M’, then A’M’ = s’ – a’.

(2) Construct the in-circle of $\triangle ABC$ so that it touches AB at M. To make things a bit easier, I assume the left circle ( in red) is larger than the right circle. Then, CD is on the right side of the median CM.

(3) From theorem (1), after some algebra, we will get IM = PK.

(4) DG = PG – PD = PI – PK = PI – IM = PM

(5) Result follows from the fact that 2PM = AP – BP