Question

answer attempt 1 out of 3
ad || bc
ab ≅ dc
m∠a =
m∠b =
m∠d =
dc =

Explanation:

Step1: Identify the quadrilateral type

The quadrilateral has \(AD \parallel BC\) and \(AB \cong DC\), so it is an isosceles trapezoid. In an isosceles trapezoid, base angles are equal, and consecutive angles between bases are supplementary.

Step2: Find \(m\angle A\)

Since \(AD \parallel BC\), \(\angle A\) and \(\angle B\) are supplementary? Wait, no, \(\angle A\) and \(\angle D\)? Wait, no, in trapezoid \(AD \parallel BC\), so \(\angle A\) and \(\angle B\)? Wait, no, let's correct. In isosceles trapezoid, \(\angle A\) and \(\angle D\) are adjacent to base \(AD\), and \(\angle B\) and \(\angle C\) are adjacent to base \(BC\). Also, \(\angle A + \angle C = 180^\circ\)? Wait, no, consecutive angles between the two bases are supplementary. So \(\angle A\) and \(\angle B\) are supplementary? Wait, no, \(AD \parallel BC\), so \(\angle A + \angle B = 180^\circ\)? Wait, no, let's look at the sides. \(AB \cong DC\), so it's isosceles trapezoid, so base angles are equal. So \(\angle C = \angle B\)? No, wait, \(\angle C\) and \(\angle D\) are adjacent, \(\angle A\) and \(\angle B\) are adjacent. Wait, given \(\angle C = 73^\circ\), in isosceles trapezoid, \(\angle A + \angle C = 180^\circ\)? Wait, no, \(\angle A\) and \(\angle D\) are equal, \(\angle B\) and \(\angle C\) are equal? Wait, no, that's not right. Wait, in isosceles trapezoid, the base angles are equal. So if \(AD\) and \(BC\) are the two bases (since \(AD \parallel BC\)), then the base angles at \(A\) and \(D\) are equal, and base angles at \(B\) and \(C\) are equal. Also, consecutive angles between the bases are supplementary. So \(\angle A + \angle B = 180^\circ\), \(\angle D + \angle C = 180^\circ\). Wait, but \(\angle A = \angle D\) and \(\angle B = \angle C\)? No, that would mean \(\angle A + \angle B = 180^\circ\) and \(\angle A = \angle D\), \(\angle B = \angle C\). Wait, given \(\angle C = 73^\circ\), so \(\angle B = 73^\circ\)? No, that can't be, because then \(\angle A + 73^\circ = 180^\circ\), so \(\angle A = 107^\circ\), and \(\angle D = 107^\circ\), \(\angle C = 73^\circ\), \(\angle B = 73^\circ\). Wait, let's check: in trapezoid \(AD \parallel BC\), so \(\angle A\) and \(\angle B\) are same - side interior angles, so they are supplementary. Similarly, \(\angle D\) and \(\angle C\) are same - side interior angles, so they are supplementary. And since \(AB \cong DC\), it's isosceles, so \(\angle A=\angle D\) and \(\angle B = \angle C\). Wait, no, that's the mistake. In isosceles trapezoid, the base angles are equal. So the angles adjacent to each leg are equal. So leg \(AB\) has angles \(\angle A\) and \(\angle B\), leg \(DC\) has angles \(\angle D\) and \(\angle C\). Since \(AB \cong DC\), \(\angle A=\angle D\) and \(\angle B=\angle C\). Wait, but if \(AD \parallel BC\), then \(\angle A + \angle B = 180^\circ\) (same - side interior angles) and \(\angle D+\angle C = 180^\circ\) (same - side interior angles). If \(\angle A=\angle D\) and \(\angle B = \angle C\), then \(\angle A+\angle B = 180^\circ\) and \(\angle A+\angle B = 180^\circ\), which is consistent. Given \(\angle C = 73^\circ\), so \(\angle B = 73^\circ\). Then \(\angle A = 180^\circ-\angle B=180 - 73=107^\circ\).

Step3: Find \(m\angle B\)

Since it's isosceles trapezoid with \(AB \cong DC\) and \(AD \parallel BC\), \(\angle B=\angle C = 73^\circ\)? Wait, no, earlier we thought \(\angle A + \angle B = 180^\circ\), but if \(\angle B=\angle C = 73^\circ\), then \(\angle A = 107^\circ\), and \(\angle D=\angle A = 107^\circ\), and \(\angle D+\angle C=107 + 73 = 180^\circ\), which is supplementary, so that works. So \(\angle B=\angle C = 73^\circ\)? Wait, no, that contradicts the supplementary. Wait, no, I think I mixed up the angles. Let's re - establish:

In trapezoid \(AD\parallel BC\), the legs are \(AB\) and \(DC\) (since \(AD\) and \(BC\) are the two bases). So the angles at \(A\) and \(D\) are on base \(AD\), angles at \(B\) and \(C\) are on base \(BC\). In isosceles trapezoid (\(AB\cong DC\)), \(\angle A=\angle D\) and \(\angle B=\angle C\). Also, since \(AD\parallel BC\), \(\angle A+\angle B = 180^\circ\) (same - side interior angles) and \(\angle D+\angle C = 180^\circ\) (same - side interior angles). Given \(\angle C = 73^\circ\), so \(\angle B=\angle C = 73^\circ\) (because \(AB\cong DC\), isosceles trapezoid property). Then \(\angle A=180^\circ-\angle B = 180 - 73=107^\circ\), and \(\angle D=\angle A = 107^\circ\) (isosceles trapezoid property).

Step4: Find \(m\angle D\)

Since \(\angle D=\angle A\) (isosceles trapezoid property) and \(\angle A = 107^\circ\), so \(m\angle D = 107^\circ\).

Step5: Find \(DC\)

Given \(AB = 32\) and \(AB\cong DC\) (from the problem statement \(AB\cong DC\)), so \(DC = AB=32\).

Answer:

\(m\angle A = 107^\circ\)
\(m\angle B = 73^\circ\)
\(m\angle D = 107^\circ\)
\(DC = 32\)