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Question
In a ΔABC, BD is the median to the side AC, BD is produced to E such that BD = DE.
Prove that: AE is parallel to BC.
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Solution
Given: A(ΔABC) in which BD is the median to AC.
BD is produced to E such that BD = DE,
We need to prove that AE II BC.
Construction: Join AE
Proof:
AD = DC ...[ BD is median to AC ] ...(1)
In ΔBDC and ΔADE,
BD = DE ...[ Given ]
∠BDC = ∠ADE = 90° ...[ Vertically opposite angles ]
AD = DC ...[ from(1) ]
∴ By Side-Angle-Side Criterion of congruence,
ΔBDC ≅ ΔADE
The corresponding parts of the congruent triangles are congruent.
∴ ∠EAD = ∠BCD ...[ c.p.c.t. ]
But these are alternate angles and AC is the transversal.
Thus, AE || BC.
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