A triangular window of a building is shown above. Its diagram represents a $\triangle ABC$ with $\angle A = 90°$ and $AB = AC$. Points P and R trisect AB and PQ II RS II AC.

A triangular window of a building is shown above. Its diagram represents a $\triangle ABC$ with $\angle A = 90°$ and $AB = AC$. Points P and R trisect AB and PQ II RS II AC.

Given:
- $\triangle ABC$ is a right-angled triangle with $\angle A = 90^\circ$ and $AB = AC$.
- Points $P$ and $R$ trisect $AB$, and $PQ \parallel RS \parallel AC$.

(a) Show that $\triangle BPQ \sim \triangle BAC$:

Since $PQ \parallel AC$, corresponding angles are equal:
$\angle BPQ = \angle BAC$ and $\angle BQP = \angle BCA$.
Thus, by the AA similarity criterion,
$\triangle BPQ \sim \triangle BAC$.

(b) Prove that $PQ = \frac{1}{3} AC$:

Since $P$ trisects $AB$, we have $AP = \frac{1}{3} AB$.
As $PQ \parallel AC$ and $\triangle BPQ \sim \triangle BAC$,
$ \frac{PQ}{AC} = \frac{BP}{BA} $.
Since $BP = \frac{1}{3} AB = \frac{1}{3} AC$,
$ PQ = \frac{1}{3} AC $.

(c) If $AB = 3$ m, find length $BQ$ and $BS$. Verify that $BQ = \frac{1}{2}BS$.

Since $AB = 3$ m and $P$ trisects $AB$,
$ AP = PB = \frac{3}{3} = 1 $ m.

Since $\triangle BPQ \sim \triangle BAC$,
$ \frac{BQ}{BC} = \frac{BP}{BA} = \frac{1}{3} $.
Thus, $ BQ = \frac{1}{3} BC $.

Similarly, since $\triangle BRS \sim \triangle BAC$,
$ \frac{BS}{BC} = \frac{BR}{BA} = \frac{2}{3} $.
Thus, $ BS = \frac{2}{3} BC $.

Now, verifying $BQ = \frac{1}{2}BS$:
$ BQ = \frac{1}{3} BC $ and $ BS = \frac{2}{3} BC $,
so $ \frac{BQ}{BS} = \frac{\frac{1}{3} BC}{\frac{2}{3} BC} = \frac{1}{2} $.
Hence, $ BQ = \frac{1}{2} BS $, which is verified.

OR

(c) Prove that $BR^2 + RS^2 = \frac{4}{9} BC^2$:

Since $\triangle BRS \sim \triangle BAC$,
$ \frac{BR}{BA} = \frac{2}{3} $, so $ BR = \frac{2}{3} BA $.

Also, $ \frac{RS}{AC} = \frac{2}{3} $, so $ RS = \frac{2}{3} AC $.

Since $BA = AC$, we get
$ BR = RS = \frac{2}{3} BC $.

Now,
$ BR^2 + RS^2 = \left(\frac{2}{3} BC\right)^2 + \left(\frac{2}{3} BC\right)^2 $
$ = \frac{4}{9} BC^2 + \frac{4}{9} BC^2 $
$ = \frac{4}{9} BC^2 $.

Hence, proved.

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