In the circle $ABCDE$, $EC$ is a diameter. Given that $\angle ABC = 158^{\circ}$, find $\angle ADE$.

1. Problem Description

In the circle $ABCDE$, $EC$ is a diameter. Given that $\angle ABC = 158^{\circ}$, find $\angle ADE$.

2. Solution Steps

Since $EC$ is a diameter, the inscribed angle $\angle CAE$ subtends a semicircle, and therefore $\angle CAE = 90^{\circ}$.

Also, quadrilateral $ABCE$ is cyclic. The opposite angles of a cyclic quadrilateral are supplementary. Therefore, $\angle AEC + \angle ABC = 180^{\circ}$.

So, $\angle AEC = 180^{\circ} - \angle ABC = 180^{\circ} - 158^{\circ} = 22^{\circ}$.

Since $EC$ is a diameter, $\angle EDC$ is an inscribed angle that subtends a semicircle. Therefore $\angle EDC = 90^{\circ}$.

Also, since $ABCDE$ is a cyclic pentagon, the quadrilateral $ACDE$ is cyclic. Thus, $\angle EAC + \angle EDC = 90^\circ + \angle ADC = 180^\circ - \angle AEC$.

Since $A, B, C, D, E$ lie on the circle, quadrilateral $ABCD$ is a cyclic quadrilateral. The opposite angles of a cyclic quadrilateral are supplementary.

Therefore, $\angle ADC + \angle ABC = 180^{\circ}$. So, $\angle ADC = 180^{\circ} - 158^{\circ} = 22^{\circ}$.

Consider quadrilateral $ADEB$. It is a cyclic quadrilateral. Thus, $\angle ADE + \angle ABE = 180^{\circ}$.

Also consider quadrilateral $CDEB$. It is a cyclic quadrilateral. Thus, $\angle CDE + \angle CBE = 180^{\circ}$.

We are given $\angle ABC = 158^\circ$. Then $\angle ABC + \angle AEC = 180^\circ$ implies $\angle AEC = 180^\circ - \angle ABC$ does not apply here.

Let's try another approach. Since $EC$ is a diameter, $\angle EAC = 90^{\circ}$. The angles subtended by the same arc are equal.

$\angle ADE = \angle ACE$

In triangle $ABC$, $\angle ABC = 158^\circ$. Since $ABCE$ is a cyclic quadrilateral, $\angle AEC = 180^\circ - 158^\circ = 22^\circ$.

Since $EC$ is the diameter, $\angle CDE = 90^\circ$.

Consider cyclic quadrilateral $ABCE$. $\angle AEC + \angle ABC = 180^{\circ}$, so $\angle AEC = 180 - 158 = 22^{\circ}$.

Since $EC$ is a diameter, the arc EC has 180 degrees. Then $\angle EAC = 90^{\circ}$.

The angle subtended at the center is twice the angle subtended at the circumference. Also, the sum of angles in a cyclic quadrilateral is $360^\circ$.

Since $ADE = ACE$ and we need to find $ADE$, we have $\angle ACE = \angle ADE = x$.

In $\triangle ACE$, $\angle AEC + \angle EAC + \angle ACE = 180^{\circ}$. Thus $22^{\circ} + 90^{\circ} + x = 180^{\circ}$.

$112^{\circ} + x = 180^{\circ}$, so $x = 180^{\circ} - 112^{\circ} = 68^{\circ}$.

3. Final Answer

68 degrees