Question

In the figure, given below, the medians BD and CE of a triangle ABC meet at G. Prove that:

1. ΔEGD ~ ΔCGB and
2. BG = 2GD from (i) above.

Answer 1

i. Since, BD and CE are medians

AD = DC

AE = BE

Hence, by converse of Basic Proportionality theorem,

ED || BC

In ΔEGD and ΔCGB,

∠DEG = ∠GCB   ...(Alternate angles)

∠EGD = ∠BGC   ...(Vertically opposite angles)

ΔEGD ~ ΔCGB    ...(AA similarity)

ii. Since, ΔEGD ~ ΔCGB

`(GD)/(GB) = (ED)/(BC)`    ...(1)

In ΔAED and ΔABC,

∠AED = ∠ABC   ...(Corresponding angles)

∠EAD = ∠BAC   ...(Common)

∴ ΔEAD ∼ ΔBAC  ...(AA similarity)

∴ `(ED)/(BC) = (AE)/(AB) = 1/2 `   ...(Since, E is the mid-point of AB)

`=> (ED)/(BC) = 1/2`

From (1),

`(GD)/(GB) = 1/2`

GB = 2GD