Day 20: Chase for the Angle

Geometry Level 3

In triangle $ABC$ , $BA=BC$ . The altitude $AD$ is drawn, then a point $P$ is selected on it. Let $E$ be the foot of the perpendicular of $AC$ from $P$ . $EP$ intersects $BC$ at $Q$ (both extended if necessary).

Given that $\angle ABC = 40^{\circ}$ , find the value of $\angle PQD$ .

This problem is part of the set Advent Calendar 2014 .

The answer is 20.

4 solutions

Jacob Coxon
Dec 24, 2014

$EQ$ is parallel to the angle bisector of $\angle ABC$

So by corresponding angles, $\angle PQD = \frac{1}{2}\angle ABC = 20 ^ \circ$

Nice solution Jacob! It's good to see you on Brilliant.

Michael Ng - 6 years, 5 months ago

Michael Ng
Dec 19, 2014

An angle chase works nicely: $\angle BCA = 70^{\circ} \\ \angle QPD = 70^{\circ} \\ \angle PQD = \boxed{20^{\circ}}$

Tijmen Veltman
Dec 21, 2014

Line $AD$ seems unnecessary, or is it meant as a distraction?

That is a great observation - I did not notice it before! It does act as a distraction.

Michael Ng - 6 years, 5 months ago

P is on AD so it's necessary from that point of view

TONI DeCESARE - 5 years, 11 months ago

Paola Ramírez
Jan 4, 2015

As $\triangle ABC$ is isosceles $\angle QCA=70°$ , $\angle QEC=90°$ .

$\angle QCA+\angle QEC+\angle PQD=180°$

$70°+90°+\angle PQD=180°$

$\boxed{\angle PQD=20°}$

Day 20: Chase for the Angle

The answer is 20.

4 solutions

0 pending reports