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D is the mid-point of side BC… (Class 10 Maths 2026, Q33a)

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MathsCBSE Class 10 · Standard · Board exam 2026, set 30/5/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2026: question 33(a)

Set in the CBSE Class 10 board exam of 2026, paper 30/5/1, section D, question 33(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(a) · 5 marks · Long answerTriangles

D is the mid-point of side BC of . 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 BC
(iii)

Figure in the paper: Triangle ABC with A at top, B bottom left, C bottom right. D is the midpoint of BC (tick marks on BD and DC). AD is drawn and extended below BC to G, with OD = DG (tick marks). E lies on AB and F lies on AC; BF and CE are drawn intersecting at O on AD. EF is drawn as a horizontal segment near the top, with E and F marked. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 30/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

A quadrilateral whose diagonals bisect each other is a parallelogram.

Hint 2

Use the parallel sides of OBGC inside triangles ABG and AGC, and apply BPT in each.

Hint 3

If a line divides two sides of a triangle in the same ratio, it is parallel to the third side; then use AA for part (iii).

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

  1. (i) In quadrilateral OBGC, [D is mid-point of BC] and [given]. So the diagonals BC and OG bisect each other at D. Hence OBGC is a parallelogram.

    1 mark

  2. (ii) Since OBGC is a parallelogram, and . E lies on line CO, so ; F lies on line BO, so .

    ½ mark

  3. In , with E on AB and O on AG, so by the Basic Proportionality Theorem ...(1)

    1 mark

  4. Similarly in , , so ...(2) [BPT]

    ½ mark

  5. From (1) and (2), . So in , EF divides AB and AC in the same ratio, hence [converse of BPT].

    1 mark

  6. (iii) In and : [corresponding angles, ] and [common]. Hence [AA similarity].

    1 mark

Answer: (i) OBGC is a parallelogram (diagonals bisect each other); (ii) EF ∥ BC (BPT and its converse); (iii) △AEF ~ △ABC (AA)

Where the marks are

  • (i) Diagonals OG and BC bisect each other ⇒ parallelogram (1 mark)
  • (ii) Deduces from the parallelogram (½ mark)
  • Applies BPT in △ABG: (1 mark)
  • Applies BPT in △AGC: (½ mark)
  • Equates ratios and uses converse of BPT to get EF ∥ BC (1 mark)
  • (iii) Equal corresponding angle + common angle ⇒ AA similarity (1 mark)

Revise the chapter

  • Chapter: Triangles (notes, formulas, where marks are lost and every past question)
Practise this chapterBoard-style questions from Triangles, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 30/5/1 (Standard): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.