Mathematics
In a △ ABC, 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
△ ABC with BD as median to the side AC and BD is produced to E such that BD = DE is shown below:
In △ ADE and △ BDC,
⇒ ∠ADE = ∠BDC (Vertically opposite angles are equal)
⇒ AD = DC (As BD is median to side AC)
⇒ BD = DE (Given)
∴ △ ADE ≅ △ BDC (By S.A.S. axiom).
We know that,
Corresponding parts of congruent triangles are equal.
∴ ∠EAD = ∠DCB
The above angles are alternate angles, since they are equal,
∴ AE || BC.
Hence, proved that AE is parallel to BC.
Related Questions
In the following diagrams, ABCD is a square and APB is an equilateral triangle. In each case,
(i) Prove that : △ APD ≅ △ BPC
(ii) Find the angles of △ DPC.
In the figure, given below, triangle ABC is right-angled at B. ABPQ and ACRS are squares. Prove that :
(i) △ ACQ and △ ASB are congruent.
(ii) CQ = BS.
ABCD is a parallelogram. The sides AB and AD are produced to E and F respectively such that AB = BE and AD = DF. Prove that :
△ BEC ≅ △ DCF
In the following figure, ABC is an equilateral triangle in which QP is parallel to AC. Side AC is produced upto point R so that CR = BP.
Prove that QR bisects PC.
