Equilateral Triangle

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Fahim Shahriar
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Equilateral Triangle

Unread post by Fahim Shahriar » Sun Sep 16, 2012 4:14 pm

$ABCD$ is a square. $P$ is a point inside the square.
$\angle PAB=\angle PBA=15º$

Prove that $∆CPD$ is an equilateral triangle.
Name: Fahim Shahriar Shakkhor
Notre Dame College

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*Mahi*
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Re: Equilateral Triangle

Unread post by *Mahi* » Sun Sep 16, 2012 8:11 pm

Hint:
Alternate segment theorem.
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SANZEED
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Re: Equilateral Triangle

Unread post by SANZEED » Sun Sep 16, 2012 11:37 pm

Solution outline:
First construct $\triangle BQC\cong \triangle APB$, with $Q$ inside $ABCD$. Now prove that $\triangle BPQ$ is equilateral, $CQ$ extended is perpendicular to $PB$ and bisects it.Then we can show that $CP=CB=CD$. Similarly $DP=DC$. :|
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Fahim Shahriar
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Re: Equilateral Triangle

Unread post by Fahim Shahriar » Wed Feb 20, 2013 1:30 pm

I used trigonometry.

Draw perpendiculars $PM$ & $PN$ on $AB$ & $CD$ respectively.
Let the side of the square $x$. $PM+PN=x$.

In $\triangle PBM$
$tan PBM = \frac {PM}{BM}$
$2 - \sqrt {3} = \frac {2PM}{x}$
$2 - \sqrt {3} = \frac {2x-2PN}{x}$
$PN = \frac {\sqrt {3}x}{2}$

In $\triangle PCN$
$tan PCN = \frac {PN}{\frac {x}{2}}$
$tan PCN = \sqrt {3}$
$PCN = 60°$

Similarly, we can prove $PDN=60°$ or we can prove $\triangle APD$ and $\triangle BCD$ are congruent; $PD=PC$.
Name: Fahim Shahriar Shakkhor
Notre Dame College

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