Question

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Explanation:

Step1: Find BC (AD=BC in rectangle)

$BC = AD = 9$

Step2: Find AB via Pythagorean theorem

In rectangle, $\triangle ABC$ is right-angled at $B$.
$AB = \sqrt{AC^2 - BC^2} = \sqrt{22^2 - 9^2} = \sqrt{484 - 81} = \sqrt{403} \approx 20.07$ (note: the handwritten 23.8 is incorrect)

Step3: Find BD (diagonals of rectangle are equal)

$BD = AC = 22$

Step4: Find EC (diagonals bisect each other)

$EC = \frac{AC}{2} = \frac{22}{2} = 11$

Step5: Find $m\angle ADC$ (all rectangle angles are 90°)

$m\angle ADC = 90^\circ$

Step6: Find $m\angle BAC$ (complementary to $\angle BCA$)

In $\triangle ABC$, $\angle BAC + \angle BCA = 90^\circ$, so $m\angle BAC = 90^\circ - 66^\circ = 24^\circ$

Step7: Find $m\angle CDB$ (alternate interior angles)

$\angle CDB = \angle BAC = 24^\circ$ (since $AB \parallel CD$)

Step8: Find $m\angle AEB$ (supplementary to $\angle BEC$)

Diagonals bisect each other, so $EB=EC$, $\angle EBC = \angle BCA=66^\circ$. In $\triangle BEC$, $\angle BEC = 180^\circ - 66^\circ -66^\circ=48^\circ$. Thus $m\angle AEB = 180^\circ - 48^\circ = 132^\circ$

Answer:

$BC = 9$
$AB = \sqrt{403} \approx 20.07$
$BD = 22$
$EC = 11$
$m\angle ADC = 90^\circ$
$m\angle BAC = 24^\circ$
$m\angle CDB = 24^\circ$
$m\angle AEB = 132^\circ$