Answer to the attachment......
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Answer:
.................................
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Answer:
in the given figure , ∠A = ∠B
Therefore ΔABC is an isosceles triangle
BC = CA
BE + EC = AD + DC
AD + EC = AD + DC { Given that AD = BE , Cancel AD from both the sides }
DC = EC
DC / EC = 1 - - - - ( I )
AD = BE (GIVEN)
AD/BE = 1 - - - - - - - (II)
From eq. I and II we get , DC/EC = AD/EB
DC / AD = EC/EB
Hence , by converse of BPT , we can say that DE ║ AB
Hence proved !!
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