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 4 of 5: Solve for the possible distances

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4x^2-4hx+ac=0\quad\Longrightarrow\quad x=\dfrac{h\pm\sqrt{h^2-ac}}{2}

Detailed analysis

Cross-multiplying the similarity relation gives $4x(h-x)=ac$ , or $4x^2-4hx+ac=0$ . The quadratic formula yields $x=(h\pm\sqrt{h^2-ac})/2$ . The distance from $X$ to $CD$ is correspondingly $h-x=(h\mp\sqrt{h^2-ac})/2$ .