geometry question-semicircle inscribed in quadrilateral. there is a line AD whose midpoint is O (AO=OD). a semicircle is drawn with centre O and any radius < AO with its straight edge being part of line AD. lines AB and CD are drawn tangent to the semicircle's arc such that BC is also a tangent to the arc. the question is to prove that $AO^2=AB·CD$
 A: I have made minor changes with suitable labels making presentation easier to read. (See below.)

Adopting the result of the special case shown earlier, we have $OA^2 = AB’. XD$.
i. e. $AO^2 = x.(t + z + x) = xt + xz + x^2$ … (1)
The target is to have $AO^2 = AB.CD$
i. e. $AO^2 = (x + y).(z + x) = xz + yz + x^2 + xy $  … (2)
Comparing the two, our target becomes "is $xt = y(x + z)$"
i. e. $\frac {x}{y} = \frac {x + z}{t}$
i. e. $\frac {x + y}{y} = \frac {x + z + t}{t}$
Construction:- (1) Join B’C’. (2) Join X O. (3) Through C, draw CE // DA cutting XO at G.
Now, by similar triangles, $\frac {x + z + t}{t} = \frac {XO}{XG}$.
There are many triangles that are similar to ⊿BAH and ⊿BB’ K. Examples are ⊿XEG and ⊿XAO.
From them, I am quite sure that $\frac {x + y}{y} = … = \frac {XO}{XG}$
That completes the proof. 
Remark: I am also sure that there are more elegant ways of proving this fact.
A: 
I hope my figure was drawn correctly according to your question. In addition, I have added the red and green circles to show the exact locations of B and T, the points of contacts of the tangents AC and DC. 
It is quite obvious that the objects are symmetric about the y-axis.
$AO^2 = DO^2$
$= CD^2 – CO^2$ … [Pythagoras theorem]
$= CD^2 – CD.CT$ … [power of a point]
$= CD.(CD – CT)$
$= CD.(CA - CB)$
$= CD.AB$
A: Triangles with the same color are congruent, and at the point $O$ we have:
$2α+2ß+2\gamma=180°$
$\alpha+\beta+\gamma=90°$
In the right-Angled triangle $OMB$, we have:
$\angle OBM=\alpha+\gamma$
Therefore the two triangles $CDO$ and $OAB$ are similar:
$\dfrac{CD}{OD}=\dfrac{OA}{AB}$
$OA\times OD=AB\times CD$
Since $OA=OD$
Thus: $OA^2=AB \times CD$

