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

In $△ A BC$ with $∠ A = 90°$ and $A B = A C$ , points $P$ and $R$ trisect $A B$ , and $PQ ∥ RS ∥ A C$ . (i) Show $△ BPQ \sim △ B A C$ by AA similarity. (ii) Prove $PQ = \frac{1}{3} A C$ using proportional sides. (iii)(a) If $A B = 3$ m, then $BP = 1$ m, and since $BP = PQ$ , $PQ = 1$ m and $BQ = 2$ m, verifying $BQ = \frac{1}{2} BS$ . OR (iii)(b) Prove $B R^{2} + R S^{2} = \frac{4}{9} B C^{2}$ using similarity ratios.

Step-by-step working

1
(i) In $△ B A C$ and $△ BPQ$ , $PQ ∥ A C$ implies $∠ BQP = ∠ BC A$ (corresponding angles) and $∠ B$ is common. Hence $△ BPQ \sim △ B A C$ by AA similarity.
1 mark
2
(ii) Since $△ BPQ \sim △ B A C$ , $\frac{PQ}{A C} = \frac{BP}{B A} = \frac{1}{3}$ (as $P$ trisects $A B$ ). Thus $PQ = \frac{1}{3} A C$ .
1 mark
3
(iii)(a) If $A B = A C = 3$ m, then $BP = 1$ m. Since $\frac{BP}{B A} = \frac{PQ}{A C}$ , $BP = PQ = 1$ m. In right $△ BPQ$ , $BQ = B P^{2} + P Q^{2} = 1 + 1 = 2$ m.
1 mark
4
Similarly, $BS = 2 BQ$ (since $S$ corresponds to $R$ which is at $\frac{2}{3} A B$ ), so $BQ = \frac{1}{2} BS = \frac{2}{2} \times 2 = 2$ m. Verified.
1 mark

Concepts used

trianglessimilar trianglesprovetrisectparallel linesfind lengthverify

Approach

Step-by-step working

Concepts used

Similar question types