Mathematics — CBSE Class 10 board question

In △BLC and △ELC... (correcting: △BLA and △ELM — let us set up correctly.)

Given: ABCD is a parallelogram. M is the midpoint of CD. BE intersects AC at L.

To Prove: EL = 2BL

Proof:

In △BLA and △ELM,

Since AB ∥ CD (opposite sides of parallelogram):
$$\angle ABL = \angle MEL \quad \text{(alternate interior angles)}$$
$$\angle BLA = \angle ELM \quad \text{(vertically opposite angles)}$$

∴ △BLA ~ △ELM (AA similarity criterion)

$$\therefore \frac{BL}{EL} = \frac{AB}{EM}$$

Now, AB = CD (opposite sides of parallelogram) and M is midpoint of CD, so:
$$EM = \frac{CD}{2} = \frac{AB}{2}$$

$$\therefore \frac{BL}{EL} = \frac{AB}{AB/2} = \frac{1}{2}$$

$$\Rightarrow EL = 2BL \qquad \textbf{(Proved)}$$

Source: Ch. 6, Section 6.4 – Criteria for Similarity of Triangles (AA similarity)

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• The key setup is identifying △BLA ~ △ELM using AA similarity (alternate angles from AB ∥ CD, and vertically opposite angles at L).
• The crucial step is recognising that since M bisects CD and AB = CD (parallelogram), we get EM = AB/2, which gives the ratio BL : EL = 1 : 2.
• Examiners award marks for: correctly naming the similar triangles, stating both angle reasons, using the midpoint condition, and concluding the ratio. Don't skip any step.