Q) 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

(Q 35 – 30/3/3 – CBSE 2026 Question Paper)

Ans:

Step 1: Let’s analyse the given diagram:

Here, ABC is a triangle and AD is the median

∴ BD = CD

Now ∵ BC = BD + CD = CD + CD

∴ BC = 2 CD

∴ ……. (i)

Step 2: ∵ BY intersects AD at X, and its given that

AX : XD = 2 : 3

So if AX = 2k and XD = 3k, then AD = AX + XD = 2k + 3k = 5k

∴

∴ ………….(ii)

Step 3: From point D, let’s draw a line DP as a parallel line to BY.

Now in Δ BCY, DP ǁ BY and D is the mid point of line BC,

then by mid – point theorem, P is the mid point of side CY.

Step 4: In Δ BCY, DP ǁ BY and D and P are mid points of sides BC and CY.

∴ By BPT theorem,

∵ By equation (i),

∴

∴ BY = 2 DP ………. (iii)

Step 5:  Let’s compare Δ ADP with Δ AXY:

Here ∠ A = ∠A           (common angle)

∠ AXY = ∠ ADP         (Corresponding angles with XY ǁ DP)

∴ By AA similarity criterion, Δ ADP ~Δ AXY:

∴ …………. (iv)

Step 6: ∵ By equation (ii),  

∴   

Now, we substituting this value in equation (iv):

∵

∴

∴ 5 XY = 2 DP ……….. (v)

Step 7: Now, we have 2 eqautions:

BY = 2 DP ………. (iii)

and 5 XY = 2 DP ……….. (v)

By comparing these 2 equations, we get: BY = 5 XY

∵ BY = BX + XY

∵ BX + XY = 5 XY

∴ BX = 5 XY – XY

∴ BX = 4 XY

Hence proved !

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