Problem 4

Let

ABCD

be a convex quadrilateral with

AB=AD+BC

. Suppose that an interior point

P

has distance

h

from the line

CD

and satisfies

AP=h+AD

and

BP=h+BC

. Prove that

\frac1{\sqrt h}\ge\frac1{\sqrt{AD}}+\frac1{\sqrt{BC}}

Step 4 of 5: Add the tangent segments

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In plain words

The three adjacent tangent lengths add exactly, so reciprocal square roots add too.

CD=DE+CE

Detailed analysis

The points occur in order $D,E,C$ on the common tangent, so $CD=DE+CE$ . Taking square roots in Step 3 gives $2\sqrt{AD\,BC}=2\sqrt{h\,AD}+2\sqrt{h\,BC}$ . Divide by $2\sqrt{h\,AD\,BC}$ to obtain $\frac1{\sqrt h}=\frac1{\sqrt{AD}}+\frac1{\sqrt{BC}}$ in the extremal case.