National BDMO 2016 : Junior 8

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National BDMO 2016 : Junior 8

Post Number:#1  Unread postby dshasan » Tue Jan 10, 2017 11:39 pm

In $\bigtriangleup ABC$ , $\angle A = 20$, $\angle B = 80$, $\angle C = 80$, $BC = 12$ units. Perpendicular $BP$ is drawn on $AC$ from from $B$ which intersects $AC$ at the point $P$. $Q$ is a point on $AB$ in such a way that $QB = 6$ units. Find the value of $\angle CPQ$.
Saadman Hasan Sommow
ST. Joseph Higher Secondary School, Dhaka
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Re: National BDMO 2016 : Junior 8

Post Number:#2  Unread postby ahmedittihad » Wed Jan 11, 2017 4:55 pm

This was a nice problem. Let $C'$ be the reflection of $C$ w.r.t $BP$. Now, $\angle C'BC = 2*10$. So, $\angle QBC'=80-20=60$. As $QB=6, BC'=12$ and $\angle QBC' =60$ we see that $\triangle QBC'$ is a $30-60-90$ triangle. So, $\angle BQC'=90$. We get, $BQC'P$ is cyclic. So, $\angle CPQ= \angle BPC+\angle BPQ=\angle BPC+BC'Q=90+30=120$.
Q.E.D
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