Question

In triangle ABC , AD is a median. X is a point on AD such that AX : XD = 2:3. Ray BX intersect AC in Y. Prove that BX= 4XY.

Answer 1

Construction: Draw DP || CY such that it intersects BY at P .

Proof:

In Δ BYC, D is the mid-point of BC and DP|| CY .

So, by Converse of Mid-Point Theorem ,

=> P is the mid-point of BY and BP=PY ------(1)

In ΔXPD & ΔXYA,

∠PXD = ∠AXY (Vertically Opposite angles )

∠PDX =∠XAY (Alernate interior angles with AC||DP)

=>ΔXPD $\sim$ ΔXYA (AA )

We know ,Corresponding parts of similar triangles are proportional .

$=>\frac{XP}{XY}= \frac{XD}{XA} \\ \\ => XP=\frac{3}{2}XY ---(2) ( since XA:XD = 2:3)$

Then,we have,

LHS= BX

= BP+PX

= PY + PX ( use eq.(1))

=[PX+XY]+3/2XY (use eq.(2))

= 3/2XY +XY+3/2XY

= 4XY

Hence Proved.

Answer 2

Answer:

Step-by-step explanation: