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

$k(O;r)$ 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 $2DG=PA_1$.

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  • $\begingroup$ 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: $DG=PG-PD=\frac {PB-\require{cancel} \cancel{BC}-AP+\cancel{AC}}{2}=\frac {PB-AP}{2}$. $\endgroup$ – farruhota Apr 21 at 16:32
  • $\begingroup$ Yes, I got it later. Thank you! $\endgroup$ – Nikol Dimitrova Apr 21 at 16:34
  • $\begingroup$ Look for similar shapes to prove the DP:AP and GP:BP are of the same proportions. $\endgroup$ – Doug M Apr 21 at 16:47
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enter image description here

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}

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  • $\begingroup$ +1. Nice drawing and a little too advanced answer. $\endgroup$ – farruhota Apr 21 at 16:47
  • $\begingroup$ What app did you use for the picture? $\endgroup$ – Dr. Mathva Apr 22 at 18:01
  • $\begingroup$ @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. $\endgroup$ – g.kov Apr 22 at 18:58
  • $\begingroup$ Thanks @g.kov! The pictures are awesome! $\endgroup$ – Dr. Mathva Apr 22 at 19:01
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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 $$PG= PK = {PB+PC-BC\over 2}$$

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  • $\begingroup$ I know this theorem but I don't know how to use it in this situation. $\endgroup$ – Nikol Dimitrova Apr 21 at 15:03
  • $\begingroup$ Mark a touching points for a start, then write some obviously equations, like $PG = PA_1$... $\endgroup$ – Aqua Apr 21 at 15:04
  • $\begingroup$ But you mean that I have to mark the other touching points because in my drawing there's only $PG = PA_1$. $\endgroup$ – Nikol Dimitrova Apr 21 at 15:08
  • $\begingroup$ 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. $\endgroup$ – Aqua Apr 21 at 15:08
  • $\begingroup$ Sorry. My English is not good. $\endgroup$ – Nikol Dimitrova Apr 21 at 15:09
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(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.

enter image description here

(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

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