Question 30
The triangle $ABC$ has $\angle BAC = 90^\circ$ and $\angle ABC = 60^\circ$. The points $D$ and $E$ are taken on the sides $AC$, respectively $AB$, so that $CD = 2 \cdot DA$ and $DE$ is the bisector of the angle $\angle ADB$. Denote $M$ the intersection of the lines $CE$ and $BD$, and $P$ the intersection of the lines $DE$ and $AM$. Prove that:
a) the lines $AM$ and $BD$ are perpendicular;
b) $3 \cdot PB = 2 \cdot CM$.
a) the lines $AM$ and $BD$ are perpendicular;
b) $3 \cdot PB = 2 \cdot CM$.