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Question
ΔABC is isosceles with AB = AC. BD and CE are two medians of the triangle. Prove that BD = CE.
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Solution
CE is median to AB
⇒ AE = BE ......(i)
BD is median to AC
⇒ AD = DC ......(i)
But AB =AC ......(iii)
Therefore from (i), (ii) and (iii)
BE = CD
In ΔBEC and ΔBDC
BE = CD
∠EBC = ∠DCB ...(angles opposites to equal sides are equal)
BC = BC ...(common)
Therefore, ΔBEC ≅ ΔBDC ...(SAS criteria)
Hence, BD = CE.
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