Question

D is the mid-point of side BC of ΔABC. CE and BF intersect at O, a point on AD. AD is produced to G such that OD = DG. Prove that

1. OBGC is a parallelogram
2. EF || BC
3. ΔAEF ∼ ΔABC

Answer 1

Given: In ΔABC, D is the mid-point of BC. O is a point on AD such that OD = DG. CE and BF intersect at O.

i. To prove OBGC is a parallelogram:

In quadrilateral OBGC, the diagonals are BC and OG.

D is the mid-point of BC   ...(Given)

D is the mid-point of OG   ...(Since OD = DG is given)

Since the diagonals BC and OG bisect each other at D, OBGC is a parallelogram.

ii. To prove EF || BC:

In ΔABG, OF is produced to B and OE is produced to C.

Since OBGC is a parallelogram:

GC || OB and BG || OC

In ΔABG, OF || BG (as BF is a part of BG line and O lies on BF)

By Basic Proportionality Theorem in ΔABG:

`(AF)/(FB) = (AO)/(OG)`

In ΔACG, OE || GC (as CE is a part of CG line and O lies on CE)

By Basic Proportionality Theorem in ΔACG:

`(AE)/(EC) = (AO)/(OG)`

Comparing both equations:

`(AF)/(FB) = (AE)/(EC)`

By Converse of Basic Proportionality Theorem in ΔABC, since the sides are divided in the same ratio:

EF || BC

iii. To prove ΔAEF ∼ ΔABC:

In ΔAEF and ΔABC:

∠A = ∠A   ...(Common angle)

∠AEF = ∠ABC   ...(Corresponding angles as EF || BC)

∠AFE = ∠ACB   ...(Corresponding angles as EF || BC)

By AA similarity criterion:

ΔAEF ∼ ΔABC