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### National BDMO 2016 : Junior 8

Posted: **Tue Jan 10, 2017 11:39 pm**

by **dshasan**

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 $\angle CPQ$.

### Re: National BDMO 2016 : Junior 8

Posted: **Wed Jan 11, 2017 4:55 pm**

by **ahmedittihad**

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

### Re: National BDMO 2016 : Junior 8

Posted: **Thu Jan 19, 2017 2:13 am**

by **Kazi_Zareer**

### Re: National BDMO 2016 : Junior 8

Posted: **Thu Feb 02, 2017 7:11 pm**

by **Thamim Zahin**

I had made the reflection but didn't get that it was a right triangle. Have to draw diagram as scale from now on.