ONMAS 2006

Solution 1:

Centrate in $\triangle APC$ that is rectangle in $P$ , we know that $D$ is middle point of $AC$ because $\triangle ACB\sim \triangle ECD$ so we deduce that $PD$ is radii of $\triangle APC$ circumcircle $\therefore \triangle PDC$ is isosceles $\Rightarrow \angle PDC=180°-40°=\boxed{140°}$

Solution 2:

Draw the heigth of $\triangle EDC$ which cut $BC$ in $K$ , then $\triangle APC\sim \triangle DKC \Rightarrow PC/KC=2 \therefore PC=2KC \Rightarrow PK=KC$ so $PDC$ is isosceles and $\angle PDC=180°-40°=\boxed{140°}$

APC is a Right- Angled Triangle.

If there is a line PD drawn from P bisecting AC then ∠APD = ∠PAC and ∠DPC =∠ACP = 20

∠PDC = 180 - ∠DPC - ∠DCP = 180 - 20 - 20 = 140