In a pentagon ABCDE, AB = AE, BC = ED and BD = EC. Prove that AC = AD.
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To prove that AC = AD in pentagon ABCDE, we can use congruence of triangles. Let's break down the steps:
1. Given: In pentagon ABCDE, AB = AE, BC = ED, and BD = EC.
2. Draw diagonal BE to split the pentagon into two triangles: ABE and CDE.
3. In triangle ABE, we have AB = AE (given) and BE (a common side).
4. In triangle CDE, we have EC = BD (given) and BE (a common side).
5. Since AB = AE and EC = BD, we can conclude that triangle ABE is congruent to triangle CDE by the Side-Side-Side (SSS) congruence criterion, as they share three sides in the same order.
6. By congruence, the remaining sides and angles in congruent triangles are also equal. Therefore, angle ABE = angle CDE.
7. Since angles ABE and CDE are opposite angles formed by the intersection of diagonal BE, we can deduce that triangle ABC is congruent to triangle CED by the Angle-Angle (AA) congruence criterion, as they share two equal angles.
8. By congruence, we can conclude that AC = AD, as the corresponding sides of congruent triangles are equal.
Thus, we have proven that AC = AD in pentagon ABCDE using congruence of triangles.