$ABCD$ is a square where $AB=4.$ $P$ is a point inside the square such that $\angle PAB = \angle PBA = 15^\circ.$ $E$ and $F$ are the midpoints of $AD$ and $BC$ respectively. $EF$ intersects $PD$ and $PC$ at points $M$ and $N$ respectively.
$Q$ is a point inside the quadrilateral $MNCD$ such that $\angle MQN = 2 \angle MPN.$ The perimeter of $\triangle MNQ$ is $4$. $PQ=?$
Triangle Inside a Square
- Fahim Shahriar
- Posts:138
- Joined:Sun Dec 18, 2011 12:53 pm
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
- Raiyan Jamil
- Posts:138
- Joined:Fri Mar 29, 2013 3:49 pm
Re: Triangle Inside a Square
I think PQ=2.19615........
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