Question

A triangular window of a building is shown above. Its diagram represents a $△ A BC$ with $∠ A = 90°$ and $A B = A C$ . Points P and R trisect AB and $PQ ∥ RS ∥ A C$ . Based on the above, answer the following questions:

(i) Show that $△ BPQ \sim △ B A C$ .

(ii) Prove that $PQ = \frac{1}{3} A C$ .

Approach

Triangular window with $∠ A = 90°$ , AB = AC = 3 m. Points P, R on AB and PQ parallel to AC. (i) Since PQ parallel AC and angle A = 90°, triangle BPQ similar to triangle BAC by AA. (ii) From similarity, $\frac{PQ}{A C} = \frac{BP}{B A} = \frac{1}{3}$ , so $PQ = \frac{1}{3} A C = 1$ m. (iii)(a) Given AB = 3, $\frac{BP}{B A} = \frac{1}{3}$ , so BP = 1 m, hence BQ = $B P^{2} + P Q^{2} = 2$ m. Since R trisects AB at $\frac{2}{3}$ position, BS = $\frac{2}{3} \times 3 = 2$ m (where S on AC). Verify $BQ = \frac{1}{2} BS \Rightarrow 2 = \frac{1}{2} \times 22$ (using similar approach). OR (iii)(b) $B R^{2} + R S^{2} = (\frac{2}{3} A B)^{2} + (\frac{2}{3} A C)^{2} = \frac{4}{9} (A B^{2} + A C^{2}) = \frac{4}{9} B C^{2}$ .

Step-by-step working

1
(i) In $△ B A C$ and $△ BPQ$ , PQ $∥$ AC $\Rightarrow ∠ BQP = ∠ BC A$ and $∠ B$ common. Hence $△ BPQ \sim △ B A C$ by AA similarity.
1 mark
2
(ii) From similarity, $\frac{PQ}{A C} = \frac{BP}{B A} = \frac{1}{3} \Rightarrow PQ = \frac{1}{3} \times 3 = 1$ m.
1 mark
3
(iii)(a) BP = 1 m, PQ = 1 m $\Rightarrow$ BQ = $1^{2} + 1^{2} = 2$ m. BS (where S on AC at $\frac{2}{3}$ from A) = $22$ m. Verify $BQ = \frac{1}{2} BS$ .
1.5 marks
4
(iii)(b) $B R^{2} + R S^{2} = (\frac{2}{3} A B)^{2} + (\frac{2}{3} A C)^{2} = \frac{4}{9} (A B^{2} + A C^{2}) = \frac{4}{9} B C^{2}$ .
1.5 marks

Concepts used

trianglessimilar trianglesparallel linesproveshow that

Approach

Step-by-step working

Concepts used

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