Question

isosceles trapezoid abcd is shown. what is the measure, in degrees, of ∠cda?

Explanation:

Step1: Recall property of isosceles trapezoid

In an isosceles trapezoid, consecutive angles between the bases are supplementary (sum to \(180^\circ\)). Here, \(AB\) and \(CD\) are the legs, \(AD\) and \(BC\) are the bases. So \(\angle ABC\) and \(\angle CDA\) are related? Wait, no: \(\angle ABC\) and \(\angle BAD\) are supplementary? Wait, no, let's correct. In trapezoid \(ABCD\) with \(AD \parallel BC\) (since it's a trapezoid, bases are parallel), so \(\angle ABC\) and \(\angle BAD\) are same - side interior angles, supplementary. But also, in isosceles trapezoid, base angles are equal. Wait, the legs are \(AB\) and \(CD\) (marked equal). So the angles adjacent to each base are equal. So \(\angle A\) and \(\angle D\) are equal, \(\angle B\) and \(\angle C\) are equal. Also, \(\angle A + \angle B= 180^\circ\) (since \(AD \parallel BC\), same - side interior angles). Given \(\angle B = 113^\circ\), so \(\angle A=180 - 113 = 67\)? Wait, no, maybe I mixed up the angles. Wait, the angle at \(B\) is \(113^\circ\), and we need \(\angle CDA\) (angle at \(D\)). Since \(AD\) is a base, and \(AB\) and \(CD\) are legs. In isosceles trapezoid, \(\angle A=\angle D\) and \(\angle B = \angle C\), and \(\angle A+\angle B = 180^\circ\) (because \(AD\parallel BC\), consecutive angles between the bases are supplementary). So \(\angle A=180^\circ-\angle B\). Since \(\angle D=\angle A\) (base angles of isosceles trapezoid), so \(\angle CDA=\angle A = 180 - 113\).

Step2: Calculate the measure of \(\angle CDA\)

\(\angle CDA=180^\circ - 113^\circ=67^\circ\)? Wait, but the initial input had 53, maybe I made a mistake. Wait, no, let's re - check. Wait, maybe the trapezoid is labeled differently. Wait, the trapezoid is \(ABCD\) with \(B\) and \(C\) on the top base, \(A\) and \(D\) on the bottom base. So \(BC\parallel AD\). Then \(\angle B\) and \(\angle A\) are same - side interior angles, so \(\angle B+\angle A = 180^\circ\). \(\angle A\) and \(\angle D\) are base angles, so \(\angle A=\angle D\). So \(\angle D = 180 - 113=67\). But the user's input had 53, maybe a typo? Wait, no, maybe I misread the angle. Wait, the angle at \(B\) is \(113^\circ\), so \(180 - 113 = 67\). So the measure of \(\angle CDA\) is \(67\). Wait, but the initial text in the box was 53, maybe that's incorrect. Let's do the calculation again. If two parallel lines are cut by a transversal, same - side interior angles are supplementary. So in trapezoid \(ABCD\), \(AD\parallel BC\), \(AB\) is a transversal, so \(\angle ABC+\angle BAD = 180^\circ\). Since it's isosceles, \(AB = CD\), so \(\angle BAD=\angle CDA\). So \(\angle CDA=180 - 113 = 67\).

Answer:

\(67\)