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

Given: D is mid-point of BC; BF and CE meet at O on AD; AD produced to G such that OD = DG.

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(i) OBGC is a parallelogram:

In quadrilateral OBGC, D is the mid-point of BC (given) and OD = DG (given), so the diagonals OG and BC bisect each other at D.
∴ OBGC is a parallelogram. (A quadrilateral whose diagonals bisect each other is a parallelogram.)

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(ii) EF ∥ BC:

Since OBGC is a parallelogram, OB ∥ GC and OC ∥ GB.
i.e., FB ∥ GC ⟹ in △ADC, F is on AC and O is on AD with FO ∥ DC (since OB ∥ GC means OC ∥ BG, so in △ABG, O on AG and F on AB give OF ∥ BG).

In △ABG: OB ∥ GC (opp. sides of parallelogram), so in △ABG, F lies on AB and O lies on AG with OF ∥ BG.
By BPT: $\dfrac{AF}{FB} = \dfrac{AO}{OG}$ … (1)

In △ACG: OC ∥ BG, so E on AC and O on AG give OE ∥ GC.
By BPT: $\dfrac{AE}{EC} = \dfrac{AO}{OG}$ … (2)

From (1) & (2): $\dfrac{AF}{FB} = \dfrac{AE}{EC}$
∴ By converse of BPT, EF ∥ BC.

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(iii) △AEF ~ △ABC:

Since EF ∥ BC,
∠AEF = ∠ABC (corresponding angles)
∠AFE = ∠ACB (corresponding angles)
∠A = ∠A (common)

∴ By AAA similarity criterion, △AEF ~ △ABC.

Source: Chapter 6, Sections 6.3 and 6.4

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• Part (i): The key property used is: if diagonals of a quadrilateral bisect each other, it is a parallelogram. D is mid-point of BC and OD = DG makes D the mid-point of OG too — so both diagonals bisect each other.
• Part (ii): Use BPT (Thales' Theorem) in triangles ABG and ACG separately, using the parallel sides of the parallelogram. Equating the ratios gives EF ∥ BC by the converse of BPT.
• Part (iii): Once EF ∥ BC is established, corresponding angles are equal → AAA criterion confirms similarity. Always write the vertices in correct correspondence: A↔A, E↔B, F↔C.