In the given figure, △ FEC ≅ △ GDB and ∠ 1 = ∠ 2. Prove that △ ADE ~ △ ABC.

△ ADE ~ △ ABC (by AA similarity), hence proved.

MathsCBSE Class 10 · Standard · Board exam 2024, set 30/5/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2024: question 34(a)

Set in the CBSE Class 10 board exam of 2024, paper 30/5/1, section D, question 34(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34(a) · 5 marks · Long answerTriangles

In the given figure, $△ F E C ≅ △ G D B$ and $∠1 = ∠2$ . Prove that $△ A D E \sim △ A B C$ .

Figure in the paper: Triangle ABC with apex A at top. D lies on AB and E lies on AC, with DE joined. Angles 1 (at D, between AD and DE) and 2 (at E, between DE and AE) are marked. Line FG is the base line; B and C lie on it between F and G. Segments FE and GD cross the figure; F is left of B, G is right of C, and FE and GD are drawn as lines from F to E and from G to D. Angles 3 and 4 are marked at B and C. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 30/5/1.

Hints

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

Hint 1

From the congruent triangles, which angles at B and C must be equal?

Hint 2

Equal angles in a triangle give equal opposite sides — apply this in both $△ A B C$ and $△ A D E$ .

Hint 3

Compare $A D / A B$ with $A E / A C$ and use the converse of the Basic Proportionality Theorem.

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.

$△ F E C ≅ △ G D B$ (given), with F↔G, E↔D, C↔B. So $∠ E C F = ∠ D B G$ [CPCT]. Since F, B, C, G are collinear, $∠ E C F = ∠4$ and $∠ D B G = ∠3$ . Hence $∠3 = ∠4$ .
1 mark
In $△ A B C$ , $∠3 = ∠4 \Rightarrow A B = A C$ ...(i) [sides opposite equal angles]
1 mark
In $△ A D E$ , $∠1 = ∠2 \Rightarrow A D = A E$ ...(ii) [sides opposite equal angles]
1 mark
Dividing (ii) by (i): $\frac{A D}{A B} = \frac{A E}{A C}$ , so D and E divide AB and AC in the same ratio; hence $D E ∥ B C$ [converse of BPT].
1 mark
As $D E ∥ B C$ : $∠1 = ∠3$ and $∠2 = ∠4$ [corresponding angles]. Therefore $△ A D E \sim △ A B C$ [AA similarity].
1 mark

Answer: $△ A D E \sim △ A B C$ (by AA similarity), hence proved.

Where the marks are

Uses congruence (CPCT) to get $∠3 = ∠4$ (1 mark)
Deduces AB = AC (1 mark)
Deduces AD = AE from $∠1 = ∠2$ (1 mark)
Gets $A D / A B = A E / A C$ and $D E ∥ B C$ (1 mark)
Corresponding angles equal and concludes 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 paper2024 board paper, set 30/5/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2024: question 34(a)

Hints

Step-by-step solution

Revise the chapter

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