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In this image

Triangle

$P$ is the point where $\overline{BY}$ and $\overline{CZ}$ cross. $\Delta ABC$ is isosceles, and proving that $\Delta BPC$ is isosceles will be enough to show that $\overline{PY}=\overline{PZ}$, but I cannot determine a way to show that $\Delta BPC$ is isosceles.

Charles
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  • information is not sufficient. is $BY$ is perpendicular to $AC$ ? –  Sep 10 '15 at 18:53
  • Not necessarily. If it's not then could it be proven that PY is not necessarily equal to PZ? I'm confident that PY=PZ, but cannot prove it. – Charles Sep 10 '15 at 19:04

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This is not true in general.

There can be two points $Y_1,Y_2$ on the side $AC$ such that $BY_1=BY_2$ where $Y_1$ is nearer to $A$.

On the other hand, there can be two points $Z_1,Z_2$ on the side $AB$ such that $CZ_1=CZ_2$ where $Z_1$ is nearer to $A$.

Now choose $Y_1,Z_2$. We can have $BY_1=CZ_2$, but $PY_1=PZ_2$ does not always hold.

mathlove
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    Yes, I encountered this while trying. thank you. My daily vote limit reached, so I cannot upvote this, sorry. –  Sep 10 '15 at 19:11
  • Is there a way to formally prove this? – Charles Sep 10 '15 at 19:43
  • @JustinD: It is sufficient to show one counterexample to disprove the statement. You can take an equilateral triangle $ABC$ to make things easier. – mathlove Sep 10 '15 at 19:46
  • @mathlove I'm not seeing how an equilateral triangle makes it easier. I can't come up with a counter example. – Charles Sep 10 '15 at 20:53
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    @JustinD: For example, let $A(1,\sqrt 3),B(0,0),C(2,0)$ and find $Y_1,Z_2$ such that $BY_1=BY_2=CZ_1=CZ_2=1.8$. Then, find $P$ and show that $PY_1\not=PZ_2$. – mathlove Sep 10 '15 at 20:58