Question

In the given figure, two triangles ABC and PQR are shown such that \(\angle A = \angle P\) and \(\angle C = \angle R\). If \(AD \perp BC\) and \(PS \perp QR\), then prove that (i) \(\Delta ADB \sim \Delta PSQ\) (ii) \(AD \times QS = BD \times PS\).

Hint:

Whenever you are asked to prove a product relation of the form \(W \times X = Y \times Z\), rearrange it as a ratio \(\frac{W}{Z} = \frac{Y}{X}\).
This immediately guides you to identify which pair of triangles you need to prove similar to get the required side ratios.

Answer 1

Correct Answer: 4

Step 1: List the given information.
We have $\Delta ABC$ and $\Delta PQR$ with $\angle A = \angle P$ and $\angle C = \angle R$. Altitudes $AD \perp BC$ and $PS \perp QR$ are drawn.
Step 2: Find the third pair of angles in the two triangles.
In $\Delta ABC$: $\angle A + \angle B + \angle C = 180^\circ$. In $\Delta PQR$: $\angle P + \angle Q + \angle R = 180^\circ$. Since $\angle A = \angle P$ and $\angle C = \angle R$, we get $\angle B = \angle Q$.
Step 3: Prove Part (i) - $\Delta ADB \sim \Delta PSQ$.
In $\Delta ADB$: $\angle ADB = 90^\circ$ (AD is altitude) and $\angle ABD = \angle B$. In $\Delta PSQ$: $\angle PSQ = 90^\circ$ (PS is altitude) and $\angle PQS = \angle Q$. Since $\angle ADB = \angle PSQ = 90^\circ$ and $\angle B = \angle Q$, by AA criterion: \[ \Delta ADB \sim \Delta PSQ \]
Step 4: Write corresponding sides from the similarity.
From $\Delta ADB \sim \Delta PSQ$: \[ \frac{AD}{PS} = \frac{DB}{SQ} = \frac{AB}{PQ} \]
Step 5: Prove Part (ii) - $AD \times QS = BD \times PS$.
From the ratio $\dfrac{AD}{PS} = \dfrac{DB}{SQ}$, cross-multiplying: \[ AD \times SQ = DB \times PS \] Since $SQ = QS$, this gives: $AD \times QS = BD \times PS$. Hence proved.
Step 6: Conclusion.
Both parts are proved using the AA similarity criterion and the proportionality of corresponding sides.
\[ \boxed{\Delta ADB \sim \Delta PSQ \text{ and } AD \times QS = BD \times PS} \]