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in Triangles by (15 points)
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In fig, If △ABE≅△ACD, show that △ADE is similar to △ABC

△ ADE is similar to  △ABC

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Given,

△ABE ≅ △ACD

Hence, AB = AC by CPCT. .....(1)

and AE = AD by CPCT.

⇒ AD = AE .....(2)

Dividing (2) by (1)

AD/AB = AE/AC ....(3)

In △ADE and △ABC

∠A = ∠A (common angle)

AD/AB = AE/AC from (3)

∴△ADE is similar to △ABC by SAS criterion.

Hence proved.

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