<s>[INST] Find the area of ​the quadrangular region $EBCD$ For reference:

Calculate the area of ​​the quadrangular region $EBCD$, if $BC = 5$, and $AD = AC$. $P$ is a tangent point. (Answer: $64$)


My progress:
$\triangle OPD \sim \triangle ADC$
$AD=2OD \implies  k=1:2$
Therefore $P$ is the midpoint $CD$ and $AP$ is perpendicular bisector of  $CD$.
Quadrilateral $ABCP$ is cyclic $(\because\angle ABC=\angle APC = 90^\circ).$
$\implies \angle PAC=\angle CBP= 26.5^\circ$
$AP$ is angle bisector.
$\triangle ACD  \implies DAC = 53^\circ$
$\therefore \angle ADC= 63.5^\circ$
$AC \parallel OP$
$BP$ is tangent to the circumference at $P$.
$\implies AC \perp BP$
$\therefore CAB = 26.5^\circ$
$BC \parallel ED (\perp AB)$
$\triangle ABC(\text{right}): (26.5^\circ, 63.5^\circ, 90^\circ) \implies(k, 2k, k\sqrt5)$
$\therefore k = 10\sqrt5 \implies AB = 2k = 10\\ BF = 2\sqrt5, BC = 5=CP=DP$
$AD = 5\sqrt5$
Any hint to finish???.....

 [/INST] From the relations you've discovered, we find $AC=5\sqrt5$ and so is $AD$.
Considering the angle $26.5^\circ$ is approximately equal to the angle we get when bisecting the second largest angle in a $3:4:5$ right triangle, we have $\tan26.5^\circ\approx\frac12$. (See also)
Therefore we can easily get $$\sin(3\cdot26.5^\circ)\approx\frac{11}{5\sqrt5}.$$ Also $$\cos(3\cdot 26.5^\circ)\approx\frac2{5\sqrt5}.$$
Now we see $\triangle AED$'s perpendicular sides are in the ratio $2:11$.
Therefore $AE=2$ and $ED=11$.
Hence the area of trapezium, $$[BCDE]=\frac{(5+11)}2\cdot8=64.$$
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