MathsCBSE Class 10 · Basic · Board exam 2025, set 430/1/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2025: question 33

Set in the CBSE Class 10 board exam of 2025, paper 430/1/1, section D, question 33, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33 · 5 marks · Long answerTriangles

State "Basic Proportionality Theorem" and use it to prove the following : In a quadrilateral ABCD, diagonals AC and BD intersect each other at O such that $\frac{A O}{B O} = \frac{C O}{D O}$ as shown in the given figure. Prove that ABCD is a trapezium.

Figure in the paper: Quadrilateral ABCD with D and C at the top, A and B at the bottom (AB is the base). Diagonals AC and BD intersect at O inside the quadrilateral. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/1/1.

Hints

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

Hint 1

Draw a line through O parallel to AB, meeting AD at E.

Hint 2

Use BPT in $△ D A B$ , rearrange the given ratio, then use the converse of BPT in $△ A D C$ .

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.

BPT: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
1 mark
Given: quadrilateral ABCD whose diagonals AC and BD meet at O with $\frac{A O}{B O} = \frac{C O}{D O}$ . To prove: $A B ∥ D C$ , i.e. ABCD is a trapezium. Construction: through O draw $O E ∥ A B$ , meeting AD at E. Figure: ABCD with diagonals meeting at O and OE parallel to AB.
1 mark
In $△ D A B$ , $O E ∥ A B$ , so $\frac{D E}{E A} = \frac{D O}{O B}$ ...(1) [by BPT]
1 mark
From the given condition, $A O \cdot D O = B O \cdot C O \Rightarrow \frac{D O}{B O} = \frac{C O}{A O}$ . With (1), $\frac{D E}{E A} = \frac{C O}{O A}$ ...(2)
½ mark
In $△ A D C$ , E lies on AD and O lies on AC with $\frac{D E}{E A} = \frac{C O}{O A}$ , so $O E ∥ D C$ [by converse of BPT].
1 mark
Now $O E ∥ A B$ and $O E ∥ D C$ , so $A B ∥ D C$ . One pair of opposite sides is parallel, so ABCD is a trapezium.
½ mark

Answer: AB ∥ DC, so ABCD is a trapezium

Where the marks are

Correct statement of BPT (1 mark)
Figure, given, to prove and construction $O E ∥ A B$ (1 mark)
Applies BPT in $△ D A B$ (1 mark)
Combines with the given ratio to get $\frac{D E}{E A} = \frac{C O}{O A}$ (½ mark)
Applies the converse of BPT in $△ A D C$ to get $O E ∥ D C$ (1 mark)
Concludes $A B ∥ D C$ , so ABCD is a trapezium (½ mark)

Revise the chapter

Chapter: Triangles (notes, formulas, where marks are lost and every past question)
Topic: Basic Proportionality Theorem (Thales Theorem)

Practise this chapterBoard-style questions from Triangles, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 430/1/1 (Basic): 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.

CBSE Class 10 Mathematics Basic 2025: question 33

Hints

Step-by-step solution

Revise the chapter

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