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Question

In ΔABC (Fig 7.60), if 1=2, prove that ABAC=BDDC.


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Solution

for triangleADB and triangleADC
angleBAD = angleCAD (given)
angleADB =angleADC ( right angle)
AD = AD (common)
Both the triangles are similar by ASA
Hence,
fraction numerator A B over denominator A C end fraction equals fraction numerator B D over denominator C D end fraction

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