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 $∥$ AC.

Based on the above, answer the following questions:

(i) Show that $△ BPQ \sim △ B A C$ .

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

(iii) (a) If $A B = 3$ m, find length BQ and BS. Verify that $BQ = \frac{1}{2} BS$ .

(iii) (b) Prove that $B R^{2} + R S^{2} = \frac{4}{9} B C^{2}$ .

Approach

A triangular window is shown with $△ A BC$ where $∠ A = 90°$ and $A B = A C$ . Points P and R trisect AB, PQ $∥$ RS $∥$ AC. (i) $△ BPQ \sim △ B A C$ by AA similarity. (ii) Since $\frac{BP}{B A} = \frac{1}{3}$ , we have $\frac{PQ}{A C} = \frac{1}{3} \Rightarrow PQ = \frac{1}{3} A C$ . (iii)(a) If $A B = 3$ m, then $BQ = 2$ m, $BS = 22$ m. Verify $BQ = \frac{1}{2} BS$ . (iii)(b) Prove $B R^{2} + R S^{2} = \frac{4}{9} B C^{2}$ using Pythagoras and similarity ratios.

Step-by-step working

1
(i) In $△ B A C$ and $△ BPQ$ , $PQ ∥ A C \Rightarrow ∠ BQP = ∠ BC A$ , $∠ B$ common. Hence $△ BPQ \sim △ B A C$ .
1 mark
2
(ii) $\frac{PQ}{A C} = \frac{BP}{B A} = \frac{1}{3} \Rightarrow PQ = \frac{1}{3} A C$ .
0.5 marks
3
(iii)(a) $BP = 1$ m, $B A = A C = 3$ m. $BQ = B P^{2} + P Q^{2} = 1 + 1 = 2$ m.
1 mark
4
$BS = 22$ m. Hence $\frac{1}{2} BS = 2 = BQ$ (verified).
0.5 marks
5
(iii)(b) $BR = \frac{2}{3} A B$ , $RS = \frac{2}{3} A C$ . $B R^{2} + R S^{2} = \frac{4}{9} (A B^{2} + A C^{2}) = \frac{4}{9} B C^{2}$ .
1 mark

Concepts used

similar trianglesbasic proportionality theoremparallel linesproveverifyfind length

Approach

Step-by-step working

Concepts used

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