Find the area of ​the quadrangular region $EBCD$

Question

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???.....

Answer 1

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, $$[BCDE]=\frac{(5+11)}2\cdot8=64.$$

Answer 2

The problem does not state that $BP$ is tangent to $\odot O$, and such a relationship is not required for the given conditions. Therefore, as stated, the problem does not have a unique solution.

However, if we do assume $BP$ is tangent to $\odot O$, then the area $$|EBCD| = \frac{25}{4} \csc^2 \frac{53\pi}{360} \left( 3 \cos \frac{37\pi}{180} - \sin \frac{7\pi}{60} \right) \approx 63.963317900822613862.$$ It cannot be exactly $64$.

If we let $\theta = 26.5^\circ = \frac{53\pi}{360}$, then it is easy to see that $\triangle APD \cong \triangle APC \cong \triangle ABC$ with $\angle BAC = \theta$, hence $\angle BAD = 3\theta$. Then $$|ABCD| = 3|AB||BC|/2 = \frac{75}{2} \cot \theta,$$ and $$r = OD = \frac{5}{2} \csc \theta,$$ and $$|\triangle AED| = r^2 \sin 6\theta.$$ Therefore, $$|EBCD| = |ABCD| - |\triangle AED|,$$ which gives the claimed result.

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As you can see, this illustrates the importance of not using approximate angle measures. If you tell me that an angle equals $26.5^\circ$, that is the exact value I will use. Mathematics is not about making unstated assumptions, or inferring that there are nice integer relationships among side lengths based on an angle measure, or claiming something is true when it is not explicitly stated.