Kaui's trapezium

First take a look at the figure below. It helps simplify a lot of things.

Alt text

Let $O$ be the center of $\Gamma$ . Join $O, A$ $;$ $O, B$ $;$ $O, C$ $;$ $O,D$ $;$ $O,E$ $;$ $O,F$ $;$ $O,G$ and $O,H$ .

Since two tangent line segments from a point outside the circle to the circle have equal lengths,

$AE=AH=2$

$BE=BF=3$

$DG=DH=x$

And $CG=CF=y$ .

Notice that $\angle BOC= \angle AOD = 90^\circ$ .

[One-liner proof: $\angle BOC= 180^\circ - (\angle OBC+\angle OCB) = 180^{\circ} - \frac{1}{2}(\angle EBC+\angle GCB) = 180^{\circ} - \frac{1}{2} \times 180^\circ= 90^\circ$ .]

Applying the Pythagoras' Theorem, we get

$AD^2=OA^2+OD^2=(AH^2+OH^2)+(OH^2+HD^2)$

$\Rightarrow (2+x)^2=(2^2+12^2)+(12^2+x^2)$

$\Rightarrow x^2+4x+4=292+x^2$

$\Rightarrow x=72$ .

Let's do the same with the right triangle $\triangle BOC$ .

That means

$BC^2=OB^2+OC^2=(BF^2+OF^2)+(OF^2+CF^2)$

$\Rightarrow (3+y)^2=(3^2+12^2)+(12^2+y^2)$

$\Rightarrow y^2+6y+9=297 + y^2$

$\Rightarrow y =48$

So, $CD=x+y=72+48=120$ .

Draw the diameter perpendicular to AB and DC: This has endpoints E and G, because both the radii are perpendicular to their respective sides, and therefore lie on this diameter. Now, drop a perpendicular from B to DC. Denote DG=x and GC=y. We obtain a right triangle with sides 12, y-3, and y+3. Therefore: $(y-3)^2+144=(y+3)^2$ , leading to y=48. Similarly, drop a perpendicular from A to DC. We obtain a right triangle with side lengths x-2, 24, and x+2, and by the Pythagorean Theorem we obtain x=72. Therefore, CD=x+y=120.

I solved it on same lines except that I dropped a perpendicular from A to CD . Let foot of perpendicular on CD be I . In triangle ADI ,DI=x-2 . Apply Pythagoras in triangle ADI . We get x. Repeat same for finding y.

It is well known that in this tangential trapezoid

$12^{4}=AE*BE*DG*GC$ and $12^{2}=\frac {5 (DG+GC)(3-DG)(2-GC)}{(DC-5)^{2}}$

Both equations yield DG+GC=120

Call the center of the circle $O$ and let $\angle HOA = \angle EOA = \alpha$ . Since perpendicular lines drawn through a point of tangency with a circle go through the center and $AB \parallel CD$ , we have $E, O, G$ collinear.

Then, we have $\angle GOD + \angle HOD + \angle HOA + \angle HOE = 180$ and $\angle GOD = \angle HOD$ , so $\angle GOD = 90 - \alpha$ . Thus, $GD = 12tan(90 - \alpha) = 12cot(\alpha) = 72$ . By a similar argument, we get $GC = 48$ , and $CD = GD + GC = 120.$