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 3 of 5: Compute tangent lengths in the extremal picture

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In plain words

A pair of right triangles turns circle tangency into product formulas.

CD^2=4AD\,BC,\quad DE^2=4h\,AD,\quad CE^2=4h\,BC

Detailed analysis

At the extremal position, let $D,C,E$ be the tangency points of $t$ with $\Gamma_A,\Gamma_B,\Gamma_P$ . Radii to tangent points are perpendicular, so $ADC$ and $BCD$ are right triangles in the limiting quadrilateral, giving $CD^2=AB^2-(AD-BC)^2=4AD\,BC$ . The standard tangent-length relation for two circles of radii $r,s$ tangent to a common line gives $DE^2=4h\,AD$ and $CE^2=4h\,BC$ .