Congruent Triangles.

Question

In the diagram, CE=CF=EF, EA=BF=2AB, and PA=QB=PC=QC=PD=QD=1, Determine BD.

Triangle APD is congruent to Triangle BQD ... So, the areas must be equal too... Please advise.

Answer 1

Note that $2 \times \angle DCB = \angle DQB$ (since $C$, $D$ and $B$ belong to the same circle with center at $Q$). But $2\times \angle DCB = \angle ACB$. The latter angle can be straightforwardly computed from $\triangle ECF$, since this triangle is equilateral. But once we know $\angle ACB$, we know $\angle DQB$, and then we can compute $BD$