In triangle ABC angle a os acute. BD and CE are perpendicular on AC and AB respectively. Prove AB x AE = AC x AD
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In △ ADB and △ AEC∠A = ∠A [Common]∠ADB = ∠AEC [Each 90°]∴ △ADB ~ △ AEC [AA − Similarity]then, ADAE = ABAC [Corresponding sides of 2 similartriangles are proportional]⇒ AB × AE = AC × AD HENCE PROVED
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In △ ADB and △ AEC∠A = ∠A [Common]∠ADB = ∠AEC [Each 90°]∴ △ADB ~ △ AEC [AA − Similarity]then, ADAE = ABAC [Corresponding sides of 2 similartriangles are proportional]⇒ AB × AE = AC × AD HENCE PROVED
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