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Rule

Based on the diagram above, the following relation holds true.

If trapezoid $ABCD$ is isosceles, then $∠A≅∠B$ and $∠C≅∠D.$

Consider an isosceles trapezoid $ABCD.$

The goal of this proof is to show that $∠A≅∠B$ and $∠C≅∠D.$

Start by drawing an auxiliary line that passes through $C$ and is parallel to $AD.$ Let $E$ be the intersection point of that line and $AB.$

Quadrilateral $AECD$ has two pairs of parallel sides, so by definition, $AECD$ is a parallelogram. Consequently, by the Parallelogram Opposite Sides Theorem, the sides opposite to each other are congruent.$AD≅EC $

By the definition of an isosceles trapezoid, $AD$ and $BC$ are congruent. Therefore, by the Transitive Property of Congruence, $BC$ and $EC$ are also congruent. $AD≅ECAD≅BC ⇒BC≅EC $

This implies that $△BEC$ is an isosceles triangle. Considering the Isosceles Triangle Theorem, it can be stated that $∠BEC$ and $∠B$ are congruent.
Additionally, $∠A$ and $∠BEC$ are congruent by the Corresponding Angles Theorem.

Therefore, using the Transitive Property of Congruence again, it is can be stated that $∠A≅∠B.$

The bases of each trapezoid are parallel and its nonparallel sides can be considered transversals. This means that $∠A$ and $∠D,$ as well as $∠B$ and $∠C,$ are consecutive interior angles.

By the Consecutive Angles Theorem, these two pairs of angles are supplementary. In other words, their corresponding sums are $180_{∘}.$${m∠A+m∠D=180_{∘}m∠B+m∠C=180_{∘} (I)(II) $

Since $∠A$ and $∠B$ are congruent, they have the same measure. Therefore, $m∠B$ can be substituted for $m∠A$ in Equation (I).
${m∠B+m∠D=180_{∘}m∠B+m∠C=180_{∘} (I)(II) $

Next, Equation (I) can be solved for $m∠D$ and Equation (II) can be solved for $m∠C.$ ${m∠D=180_{∘}−m∠Bm∠C=180_{∘}−m∠B $

The right-hand sides of the equations are the same, so $∠C$ and $∠D$ have the same measure. Therefore, $∠C$ and $∠D$ are congruent angles. This concludes the proof.
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