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Question
ABC is an isosceles triangle in which AC = BC. AD and BE are respectively two altitudes to sides BC and AC. Prove that AE = BD.
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Solution
Given: ΔABC is an isosceles triangle in which AC = BC.
Also, AD and BE are two altitudes to sides BC and AC, respectively.
To prove: AE = BD
Proof: In ΔABC,
AC = BC ...[Given]
∠ABC = ∠CAB ...[Angles opposite to equal sides are equal]
i.e., ∠ABD = ∠EAB ...(i)
In ΔAEB and ΔBDA,
∠AEB = ∠ADB = 90° ...[Given, AD ⊥ BC and BE ⊥ AC]
∠EAB = ∠ABD ...[From equation (i)]
And AB = AB ...[Common side]
∴ ΔAEB ≅ ΔBDA ...[By AAS congruence rule]
⇒ AE = BD ...[By CPCT]
Hence proved.
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