Friendly Triangles[self-made]

[Phlembac Adib Hasan]

$ABCD$ is a parallelogram. Extend $AD$ and take a point $P$ on its extension. $BP$ intersect $CD$ at $Q$.$\bigtriangleup ADQ=30$,Then $\bigtriangleup CPQ=?$

[sakibtanvir]

I think it will be 30 as well...........(Really they r friendly )

[*Mahi*]

Yup, it is.

[Phlembac Adib Hasan]

I think it will be 30 as well...........(Really they r friendly )

[Phlembac Adib Hasan]

My proof:

\[ \frac {1} {2} ABCD= \triangle AQB = \triangle ADQ+ \triangle BCQ= \triangle CPQ+ \triangle BCQ \]
\[ \Rightarrow \triangle ADQ= \triangle CPQ =30 \]