High School

Thank you for visiting Given AB cong AE BC cong DE Prove angle ACD cong angle ADC Complete the paragraph proof We are given AB cong AE and BC. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!

Given: \( AB \cong AE \); \( BC \cong DE \)
Prove: \( \angle ACD \cong \angle ADC \)

Complete the paragraph proof.

We are given \( AB \cong AE \) and \( BC \cong DE \). This means \( \triangle ABE \) is an isosceles triangle. Base angles in an isosceles triangle are congruent based on the isosceles triangle theorem, so \( \angle ABE \cong \angle AEB \). We can then determine \( \triangle ABC \cong \triangle AED \) by the Side-Angle-Side (SAS) Postulate. Because of CPCTC (Corresponding Parts of Congruent Triangles are Congruent), segment \( AC \) is congruent to segment \( AD \). Triangle \( ACD \) is an isosceles triangle based on the definition of an isosceles triangle. Therefore, based on the isosceles triangle theorem, \( \angle ACD \cong \angle ADC \).

Answer :

Final answer:

The proof establishes that triangles ABC and AED are congruent using the Angle-Side-Angle postulate, leading to the conclusion that triangle ACD is isosceles and therefore the angles ∠ACD and ∠ADC are congruent.

Explanation:

We are given that AB ≅ AE and BC ≅ DE, implying that triangle ABE is an isosceles triangle. According to the Isosceles Triangle Theorem, the base angles of an isosceles triangle are congruent, which means that ∠ABE ≅ ∠AEB. With congruent corresponding angles and sides, we can use the Angle-Side-Angle (ASA) postulate to conclude that △ABC ≅ △AED.

By the Corresponding Parts of Congruent Triangles are Congruent (CPCTC), segment AC is congruent to segment AD. Hence, triangle ACD is an isosceles triangle. Since the base angles of an isosceles triangle are congruent, we can deduce that ∠ACD ≅ ∠ADC, completing the proof.

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Rewritten by : Jeany