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In a parallelogram $ABCD$, the bisector of angle $ABC$ intersects $AD$ at $P$ where $PD=5, BP=6$ and $CP=6$. Find $AB$.

I tried equating areas and heron's formula but no luck..

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We have $$\angle{ABP}=\angle{CBP}=\angle{BPA}=\angle{BCP}$$ from which we know that $\triangle{ABP}$ and $\triangle{PBC}$ are similar.

Let $AB=x$. Then, $AP=x,BC=x+5$ to have $$AB:PB=PB:CB\iff x:6=6:x+5.$$

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  • $\begingroup$ It is obvious. But how to use it? $\endgroup$ – Roman83 May 30 '16 at 6:08
  • $\begingroup$ @Roman83: I added some details. $\endgroup$ – mathlove May 30 '16 at 6:09
  • $\begingroup$ Your solution is much better than mine. $\endgroup$ – Eclipse Sun May 30 '16 at 6:11
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Suppose $AB=x$. It is easy to see that $AB=AP$. Let $h$ be the distance between $AD$ and $BC$.

By Pythagorean theorem (try to find out the right triangles I am using), we have $$h^2+\left(\frac{x+5}2\right)^2=6^2$$ and $$h^2+\left(5-\frac{x+5}2\right)^2=x^2.$$

We have $x^2+5x-36=0$. Hence, $x=4$.

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