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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. l AD ≅ EC AD ≅ 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^(∘) & (I) m∠ B + m∠ C = 180^(∘) & (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^(∘) & (I) m∠ B + m∠ C = 180^(∘) & (II) Next, Equation (I) can be solved for m∠ D and Equation (II) can be solved for m∠ C. m∠ D = 180^(∘) - m∠ B m∠ 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.
