Mathematics — Standard (041) — CBSE Class 10 board question

Given: ABCD is a parallelogram. M is the mid-point of CD. Line BM intersects AC at L and AD produced at E.

To Prove: EL = 2BL

Proof:

In △BMC and △EMD:

• MC = MD (M is mid-point of CD)
• ∠BMC = ∠EMD (Vertically opposite angles)
• ∠BCM = ∠EDM (Alternate interior angles, BC ∥ AD)

∴ △BMC ≅ △EMD (ASA congruence)

∴ BC = ED … (1)

Now, BC = AD (Opposite sides of parallelogram) … (2)

From (1) and (2): ED = BC = AD

So, EA = ED + DA = BC + BC = 2BC … (3)

In △EAL and △CBL:

• ∠AEL = ∠BCL (Alternate interior angles, BC ∥ AE)
• ∠ELA = ∠BLC (Vertically opposite angles)

∴ △EAL ~ △CBL (AA similarity criterion)

$$\therefore \frac{EL}{BL} = \frac{EA}{CB} = \frac{2BC}{BC} = 2$$

$$\boxed{EL = 2BL}$$

Hence proved.

Source: Triangles, Section 6.4 (AA Similarity Criterion)

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Key steps examiners look for:

1. Proving △BMC ≅ △EMD to get BC = DE — this is the critical step that most students miss.
2. Establishing EA = 2BC using the parallelogram property (BC = AD).
3. Setting up △EAL ~ △CBL using AA criterion (alternate angles + vertically opposite angles).
4. Concluding EL/BL = EA/CB = 2, hence EL = 2BL.

Write the congruence step clearly before similarity — without it, the ratio EA/CB cannot be established. The ASA congruence and AA similarity criteria must be explicitly named.