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Question
AD and BE are altitudes of an isosceles triangle ABC with AC = BC. Prove that AE = BD.
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Solution
In ΔCAD and ΔCBE
CA = CB ...(Isosceles triangles)
∠CDA = ∠CEB = 90°
∠ACD = ∠BCE = ...(common)
Therefore, ΔCAD ≅ ΔCBE ...(AAS criteria)
Hence, CE = CD
But, CA = CB
⇒ AE + CE = BD + CD
⇒ AE = BD.
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