In the given figure, ∠ ABC = ∠ ACB and BC/BE = BD/AC. Show that △ ABE ~ △ DBC and AE ∥ DC.

△ ABE ~ △ DBC (SAS) and hence AE ∥ DC.

MathsCBSE Class 10 · Basic · Board exam 2024, set 430/3/1 · 3 marks · Short answer

CBSE Class 10 Mathematics Basic 2024: question 30

Set in the CBSE Class 10 board exam of 2024, paper 430/3/1, section C, question 30, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q30 · 3 marks · Short answerTriangles

In the given figure, $∠ A B C = ∠ A C B$ and $\frac{B C}{B E} = \frac{B D}{A C}$ . Show that $△ A B E \sim △ D B C$ and $A E ∥ D C$ .

Figure in the paper: Points B and C are at the base. Point D is above and to the left of the figure, with line segments from B to D and from D to C. Point A lies inside the figure, joined to B and to C. Point E lies on segment BC between B and C, and A is joined to E. Segments BD and AC are in the figure as described in the given ratio. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 430/3/1.

Hints

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

Hint 1

Equal base angles tell you two sides of $△ A B C$ are equal; replace AC by AB in the given ratio.

Hint 2

Rearrange to get a ratio of sides of $△ A B E$ to $△ D B C$ with included angle B common.

Hint 3

After similarity, look at angles at A and D with respect to line BD.

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 $△ A B C$ , $∠ A B C = ∠ A C B \Rightarrow A C = A B$ [sides opposite equal angles]. Given $\frac{B C}{B E} = \frac{B D}{A C} = \frac{B D}{A B}$ , so $\frac{A B}{D B} = \frac{B E}{B C}$ .
1 mark
In $△ A B E$ and $△ D B C$ : $\frac{A B}{D B} = \frac{B E}{B C}$ and $∠ A B E = ∠ D B C$ [common angle B]. Therefore $△ A B E \sim △ D B C$ [SAS similarity criterion].
1 mark
Corresponding angles of similar triangles are equal, so $∠ B A E = ∠ B D C$ .
½ mark
These are corresponding angles made by lines AE and DC with transversal BD, and they are equal. Hence $A E ∥ D C$ .
½ mark

Answer: $△ A B E \sim △ D B C$ (SAS) and hence $A E ∥ D C$ .

Where the marks are

Uses $A B = A C$ to get $\frac{A B}{D B} = \frac{B E}{B C}$ (1 mark)
Common $∠ B$ and SAS similarity concluded (1 mark)
$∠ B A E = ∠ B D C$ (½ mark)
Corresponding angles equal, so $A E ∥ D C$ (½ 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 430/3/1 (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 2024: question 30

Hints

Step-by-step solution

Revise the chapter

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