Question

(a) In the given figure, $△ A BC$ is right angled triangle with $∠ A = 9 0^{\circ}$ . AD is perpendicular to BC.

Prove that:

(i) $△ D B A \sim △ D A C$

(ii) $D A^{2} = D B \times D C$

(iii) Find the area of $△ A BC$ when $D B = 9$ cm and $D C = 16$ cm.

(b) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.

Approach

(a)(i): In right triangle ABC with $∠ A = 90°$ and AD perpendicular to BC, triangles DBA and DAC share angle A = 90°. Also, $∠ B D A = ∠ C D A = 90°$ (given AD ⊥ BC). Since $∠ B A D = 90° - ∠ B = ∠ A CB$ , we have $△ D B A \sim △ D A C$ by AA similarity. | (a)(ii): From $△ D B A \sim △ D A C$ , corresponding sides are proportional: $\frac{D A}{D C} = \frac{D B}{D A}$ . Cross-multiplying gives $D A^{2} = D C \times D B$ . | (a)(iii): Given DC = 16 cm and DB = 9 cm, we have $D A^{2} = 16 \times 9 = 144$ , so $D A = 12$ cm. The base BC = 16 + 9 = 25 cm. Area of triangle ABC $= \frac{1}{2} \times BC \times D A = \frac{1}{2} \times 25 \times 12 = 150$ cm². | Q35(b): To prove the Basic Proportionality Theorem: In triangle ABC, if DE is parallel to BC intersecting AB at D and AC at E, then $\frac{A D}{D B} = \frac{A E}{EC}$ . Construct perpendiculars DM to AC and EN to AB. The area of triangle ADE with base AD is $\frac{1}{2} A D \times EN$ , and with base AE is $\frac{1}{2} A E \times D M$ . Similarly, triangles DBE and DCE have equal areas (same base and between same parallels). By comparing ratios of areas, we establish the result.

Step-by-step working

1
(a)(i) In $△ D B A$ and $△ D A C$ : $∠ B D A = ∠ C D A = 90°$ (AD ⊥ BC).
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2
(a)(i) $∠ B A D = 90° - ∠ B = ∠ A CB$ . Therefore, $△ D B A \sim △ D A C$ by AA similarity.
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(a)(ii) From similarity: $\frac{D A}{D C} = \frac{D B}{D A}$ .
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(a)(ii) Cross-multiply: $D A^{2} = D C \times D B$ .
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(a)(iii) $D A^{2} = 16 \times 9 = 144 \Rightarrow D A = 12$ cm.
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(a)(iii) BC $= D C + D B = 16 + 9 = 25$ cm.
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(a)(iii) Area $= \frac{1}{2} \times 25 \times 12 = 150$ cm².
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Q35(b) Given: In $△ A BC$ , $D E ∥ BC$ . To Prove: $\frac{A D}{D B} = \frac{A E}{EC}$ .
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Q35(b) Construction: Draw $D M ⊥ A C$ and $EN ⊥ A B$ . Join BE and CD.
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Q35(b) $\frac{ar ( △ A D E )}{ar ( △ D BE )} = \frac{\frac{1}{2} A D \times EN}{\frac{1}{2} D B \times EN} = \frac{A D}{D B}$ .
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Q35(b) $\frac{ar ( △ A D E )}{ar ( △ D CE )} = \frac{\frac{1}{2} A E \times D M}{\frac{1}{2} EC \times D M} = \frac{A E}{EC}$ .
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Q35(b) Since $△ D BE$ and $△ D CE$ lie on same base and between same parallels, ar(DBE) = ar(DCE). Therefore, $\frac{A D}{D B} = \frac{A E}{EC}$ .
1 mark

Concepts used

similar trianglesproveright triangleareabasic proportionality theorem

Approach

Step-by-step working

Concepts used

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