$ABCD$ is a square. $P$ is a point inside the square.
$\angle PAB=\angle PBA=15º$
Prove that $∆CPD$ is an equilateral triangle.
Equilateral Triangle
- Fahim Shahriar
- Posts:138
- Joined:Sun Dec 18, 2011 12:53 pm
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
Re: Equilateral Triangle
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Re: Equilateral Triangle
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- Fahim Shahriar
- Posts:138
- Joined:Sun Dec 18, 2011 12:53 pm
Re: Equilateral Triangle
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; $PD=PC$.
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
Notre Dame College