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 3 of 5: Use similar triangles on either side of X

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\triangle LBX\sim\triangle MXC\quad\Longrightarrow\quad \frac{2x}{a}=\frac{c}{2(h-x)}

Detailed analysis

At a right-angle point $X$ , the angles in the small triangles $LBX$ and $MXC$ correspond, so these triangles are similar. Since $LB=a/2$ and $MC=c/2$ , similarity gives $LX/MC=LB/MX$ , namely $2x/a=c/[2(h-x)]$ .