# Two circles inscribed in isosceles triangle

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

$$k(O;r)$$ 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 $$CP$$.

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

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

• Usually in solving problems you are required to use all the given conditions. In the discussion of the first deleted answer you were one step to finish: $DG=PG-PD=\frac {PB-\require{cancel} \cancel{BC}-AP+\cancel{AC}}{2}=\frac {PB-AP}{2}$. – farruhota Apr 21 at 16:32
• Yes, I got it later. Thank you! – Nikol Dimitrova Apr 21 at 16:34
• Look for similar shapes to prove the DP:AP and GP:BP are of the same proportions. – Doug M Apr 21 at 16:47

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}

• +1. Nice drawing and a little too advanced answer. – farruhota Apr 21 at 16:47
• What app did you use for the picture? – Dr. Mathva Apr 22 at 18:01
• @Dr. Mathva: Asymptote, "Asymptote is a powerful descriptive vector graphics language that provides a natural coordinate-based framework for technical drawing. Labels and equations are typeset with LaTeX, for high-quality PostScript output." Also, you can check out this forum. – g.kov Apr 22 at 18:58
• Thanks @g.kov! The pictures are awesome! – Dr. Mathva Apr 22 at 19:01

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 $$PG= PK = {PB+PC-BC\over 2}$$

• I know this theorem but I don't know how to use it in this situation. – Nikol Dimitrova Apr 21 at 15:03
• Mark a touching points for a start, then write some obviously equations, like $PG = PA_1$... – Aqua Apr 21 at 15:04
• But you mean that I have to mark the other touching points because in my drawing there's only $PG = PA_1$. – Nikol Dimitrova Apr 21 at 15:08
• Is that so hard? You don't have to find them, they are there. Just mark them. And put another picture when you do that. – Aqua Apr 21 at 15:08
• Sorry. My English is not good. – Nikol Dimitrova Apr 21 at 15:09

(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