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.

Answer 1

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.