In the given figure, DE ∥ AC and BE/EC = BC/CP. Prove that DC ∥ AP.

BD/DA = BC/CP, so DC ∥ AP by the converse of BPT.

MathsCBSE Class 10 · Basic · Board exam 2023, set 430/2/1 · 2 marks · Very short answer

CBSE Class 10 Mathematics Basic 2023: question 21

Set in the CBSE Class 10 board exam of 2023, paper 430/2/1, section B, question 21, for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q21 · 2 marks · Very short answerTriangles

In the given figure, $D E ∥ A C$ and $\frac{B E}{E C} = \frac{B C}{C P}$ . Prove that $D C ∥ A P$ .

Figure in the paper: Triangle ABP with A at the top. Point D lies on side AB. Points B, E, C, P lie in order on the base line BP. Segment DE is drawn parallel to AC (both arrows marked), and segment AC joins A to C on the base. Segment DC is drawn from D to C. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 430/2/1.

Hints

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

Hint 1

Which triangle contains the parallel lines DE and AC? Apply the Basic Proportionality Theorem there.

Hint 2

You will get a ratio BD/DA. Compare it with the given ratio BC/CP.

Hint 3

If a line divides two sides of $△ A B P$ in the same ratio, what does the converse of BPT say about it?

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.

In $△ A B C$ , $D E ∥ A C$ . By the Basic Proportionality Theorem, $\frac{B E}{E C} = \frac{B D}{D A}$ ...(1)
1 mark
It is given that $\frac{B E}{E C} = \frac{B C}{C P}$ ...(2). From (1) and (2), $\frac{B D}{D A} = \frac{B C}{C P}$ .
½ mark
In $△ A B P$ , D lies on AB and C lies on BP, and the line DC divides these two sides in the same ratio. Therefore $D C ∥ A P$ [converse of BPT]. Hence proved.
½ mark

Answer: $\frac{B D}{D A} = \frac{B C}{C P}$ , so $D C ∥ A P$ by the converse of BPT.

Where the marks are

Applies BPT in $△ A B C$ to get $\frac{B E}{E C} = \frac{B D}{D A}$ (1 mark)
Combines with the given ratio to get $\frac{B D}{D A} = \frac{B C}{C P}$ (1/2 mark)
Concludes $D C ∥ A P$ by the converse of BPT in $△ A B P$ (1/2 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 paper2023 board paper, set 430/2/1 (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 2023: question 21

Hints

Step-by-step solution

Revise the chapter

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