# ABCD is a square. The diagonals AC & BD cut at E. From B a perpendicular is drawn to the bisector of /_DCEand it cuts AC atP and DC at Q. Can you prove that DQ = 2PE?

Oct 24, 2017

see explanation.

#### Explanation: Let $A B = s , \implies A C = B D = \sqrt{2} s , \implies B E = \frac{\sqrt{2} s}{2}$,
given $C R$ is the angle bisector of $\angle D C E$ and $B Q$ is perpendicular to $C R$,
$\implies \angle Q C R = \angle P C R = {22.5}^{\circ}$
$\implies \Delta C R Q \mathmr{and} \Delta C R P$ are congruent,
$\implies \textcolor{red}{C Q = C P}$
as $B R$ is perpendicular to $C R$,
$\implies \angle C B R = 90 - 22.5 - 45 = {22.5}^{\circ}$
$\implies \angle Q B E = 45 - 22.5 = {22.5}^{\circ} = \angle C B R$,
$\implies B Q$ is the angle bisector of $\angle C B D$,
By angle bisector theorem,
in $\Delta B C E , \frac{C P}{P E} = \frac{B C}{B E} = \frac{s}{\frac{\sqrt{2} s}{2}} = \frac{2}{\sqrt{2}} - - - E Q \left(1\right)$
in $\Delta B C D , \frac{C Q}{D Q} = \frac{B C}{B D} = \frac{s}{\sqrt{2} s} = \frac{1}{\sqrt{2}} - - - E Q \left(2\right)$
$\frac{E Q \left(1\right)}{E Q \left(2\right)} , \implies \frac{C P}{P E} \cdot \frac{D Q}{C Q} = \frac{2}{\sqrt{2}} \cdot \sqrt{2} = 2$,
since $C Q = C P , \implies \frac{D Q}{P E} = 2$
Hence $D Q = 2 P E$