Question

In triangle ABC, AD is a median, X is a point on AD such that AX:XD=2:3. Ray BX intersects AC in Y. Prove that BX=4Xy.
Plzzz answer it fast

Answer 1

Use thals therum.. And solve it..

Answer 2

Step-by-step explanation:

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

Thus, we have: In Δ BYC, D is the mid-point of BC and DP is parallel to CY.

Using the Converse of Mid-Point Theorem,  

P is the mid-point of BY, which implies that: BP = PY → (1)

In ΔXPD and ΔXYA,

∠PXD = ∠AXY; since they are vertically opposite angles.

and

∠PDX = ∠XAY; since they are alternate interior angles with AC || DP

Thus, we have:

ΔXPD ~ ΔXYA  

We know that, corresponding parts of similar triangles are proportional.

Thus, we have:

$\frac{X P}{X Y}=\frac{X D}{X A}$

Since, XA:XD = 2:3, we have:

$X P=\frac{3}{2} X Y \rightarrow(2)$

Thus, we have,  

BX = BP + PX

BX = PY + PX; Using (1)

$B X=[P X+X Y]+\frac{3}{2 X Y} ; \ \text{Using}(2)$

Further solving gives us:

$B X=\frac{3}{2 X Y}+X Y+\frac{3}{2 X Y}$

Which results into:

BX = 4XY  

Hence proved.