Question

In Δ ABC, AD is a median. X is a point on AD such that AX : XD = 2: 3. BX is extended so that, it intersects AC at Y. Prove that BX = 4 XY.

 

Answer 1

Given: Δ ABC, AD is the median of triangle x is a point of AD such that AX : XD = 2 : 3

Draw: DZ || By

In Δ AXY and Δ ADZ,

∠XAY = ∠DAZ   ...(Common angle)

∠AYB = ∠AZD   ...(Corresponding angle)

Δ AXY ∼ Δ ADZ   ...(AA similarity)

`(XY)/(DZ) = (AX)/(AZ) = 2/5`

`(XY)/(DZ) = 2/5`

5 XY = 2 DZ    ...(i)

In Δ CDZ and Δ CBY,

∠DCZ = ∠BCY   ...(Common angle)

∠CZD = ∠CYB   ...(Corresponding angle)

Δ CDZ ∼ Δ CBY   ...(AA similarity)

`(BY)/(DZ) = (BC)/(DC) = 2/1`

`(BY)/(DZ) = 2/1`

BY = 2 DZ   ...(ii)

from (i) and (ii)

5 XY = BY

`(BY)/(XY) = 5/1`

`(BY)/(XY) - 1 = 5/1 - 1`

`(BY - XY)/(XY) = (5 - 1)/1`

`(BX)/(XY) = 4/1`

BX = 4 XY