Problem 7

In the isosceles trapezoid

ABCD

with

AB\parallel DC

and

BC=AD

, let

AB=a

CD=c

, and let the perpendicular distance from

A

CD

h

. Show how to construct all points

X

on the axis of symmetry such that

\angle BXC=\angle AXD=90^\circ

. Find the distance of each such

X

from

AB

and from

CD

, and give the condition for such points to exist.

Step 1 of 5: Reduce the right-angle condition using Thales' theorem

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In plain words

The two right-angle requirements are really one circle-intersection condition because of symmetry.

\angle BXC=90^\circ\iff X\text{ lies on the circle with diameter }BC

Detailed analysis

The locus of points subtending a right angle over segment $BC$ is the circle with diameter $BC$ . Thus a required point $X$ on the symmetry axis must be an intersection of that circle with the axis. Because the trapezoid is isosceles and the symmetry axis exchanges $A\leftrightarrow B$ and $D\leftrightarrow C$ , any such $X$ also satisfies $\angle AXD=90^\circ$ .