Q) D is the mid point of side BC of Δ ABC. CE and BF intersect at O, a point on AD. AD is produced to G such that OD = DG. 

Prove that:
(i) OBGC is a parallelogram.
(ii) EF ǁ BC
(iii) Δ AEF ~ Δ ABC

(Q 34 A – 30/5/2 – CBSE 2026 Question Paper)

Ans:

(i) Prove that OBGC is a parallelogram.

Step 1: Given that D is the mid point of side BC

∴ BD = DC

Given OD = DG

Step 2: We know that diagonals of a parallelogram always bisect each other

∴ OBGC is a parallelogram…. Hence Proved!

(ii) Prove that EF ǁ BC

Step 3: ∵ OBGC is a parallelogram ∴ OC ǁ BG

Since OC is extended to E ∴ OE ǁ BG

Step 4: Next in Δ ABG, OE ǁ BG

∴ By BPT or Basic Proportionality Theorem (Thales’s Theorem),

……… (i)

Step 5: Similarly BO ǁ GC (∵ OBGC is a parallelogram)

Since BO is extended to F ∴ OF ǁ GC

Step 6: Next in Δ ACG,

∵ OF ǁ GC

∴ Again by BPT Theorem,

…….. (ii)

Step 7: By comparing equation (i) and (ii), we get:

Step 8: Now by converse of the BPT, if a line divides two sides of a triangle in the same ratio, it must be parallel to the third side.

∴ EF ǁ BC …. Hence Proved!

(iii) Prove that Δ AEF ~ Δ ABC

Step 9: ∵ EF ǁ BC (proved above)

and line AB intersects it

∴ ∠ AEF = ABC (corresponding angles)

∠ EAF = ∠ BAC (common angle)

∴ By AA similarity criterion, Δ AEF ~ Δ ABC…. Hence Proved !

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