Square perpendicular bisector

Geometry Level 3

A point $P$ is chosen on a square with center $O$ such that the perpendicular bisector of $\overline{OP}$ intersects a corner of the square. What is the measure, in degrees, of the angle between the bisector and the side containing $P$ (angle $x$ below)?

Round to the nearest tenth of a degree.

The answer is 22.5.

1 solution

Abha Vishwakarma
Oct 7, 2018

Look at this figure

We know three things for certain,

$OE = OP$
$\angle AEO = \angle AEP = 90^\circ$
$AE =AE$ (well that's obvious)

So due to $Side-Angle-Side$ similarity, $\bigtriangleup AEO$ and $\bigtriangleup AEP$ are similar.

So, $\angle OAE = x$ . Since $O$ is the centre of the square, $\angle OAP = 45^\circ$ which is also equal to 2 times the angle $x$ .

Hence $x = (\frac{45}{2})^\circ = 22.5^\circ$ .