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 1 of 5: Make the first two circles tangent

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In plain words

The distance equations are exactly tangency equations for three circles.

AB=AD+BC

Detailed analysis

Draw $\Gamma_A$ centered at $A$ with radius $AD$ and $\Gamma_B$ centered at $B$ with radius $BC$ . Since $AB=AD+BC$ , the circles are externally tangent on the segment $AB$ . Draw also $\Gamma_P$ centered at $P$ with radius $h$ ; the equations $AP=h+AD$ and $BP=h+BC$ say that it is externally tangent to both $\Gamma_A$ and $\Gamma_B$ .