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(a) State and Prove "Basic Proportionality Theorem".
OR
(b) In the given figure, $CM$ and $RN$ are respectively, the medians of $\Delta ABC$ and $\Delta PQR$. If $\Delta ABC \sim \Delta PQR$, prove that :
(i) $\Delta AMC \sim \Delta PNR$
(ii) $\angle BCM = \angle QRN$
(iii) $\Delta BMC \sim \Delta QNR$
OR
(b) In the given figure, $CM$ and $RN$ are respectively, the medians of $\Delta ABC$ and $\Delta PQR$. If $\Delta ABC \sim \Delta PQR$, prove that :
(i) $\Delta AMC \sim \Delta PNR$
(ii) $\angle BCM = \angle QRN$
(iii) $\Delta BMC \sim \Delta QNR$
Show SolutionHide Solution↓
(a) Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. (Correct Statement: $1$ mark)
Given: In $\Delta ABC, DE \parallel BC$
To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$
Construction: Draw $DM \perp AC, EN \perp AB$, join $BE$ and $CD$ (Given + To prove + Construction + Figure: $1$ mark)
Proof: $\frac{ar(\Delta ADE)}{ar(\Delta DBE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \dots (i)$ [$1$ mark]
$\frac{ar(\Delta ADE)}{ar(\Delta ECD)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \dots (ii)$ [$1$ mark]
as $\Delta DBE$ and $\Delta DCE$ lie on the same base $DE$ and between same parallels $BC$ and $DE$
$\therefore ar(\Delta DBE) = ar(\Delta ECD)$ or $\frac{ar(\Delta ADE)}{ar(\Delta DBE)} = \frac{ar(\Delta ADE)}{ar(\Delta ECD)} \dots (iii)$ [$\frac{1}{2}$ mark]
From $(i), (ii)$ and $(iii)$, we get $\frac{AD}{DB} = \frac{AE}{EC}$ [$\frac{1}{2}$ mark]
OR
(b) (i) $\Delta ABC \sim \Delta PQR \Rightarrow \frac{AB}{PQ} = \frac{AC}{PR}$ [$\frac{1}{2}$ mark]
$\Rightarrow \frac{AC}{PR} = \frac{\frac{1}{2} AB}{\frac{1}{2} PQ} \Rightarrow \frac{AC}{PR} = \frac{AM}{PN}$ [$\frac{1}{2}$ mark]
Also $\angle A = \angle P$ [$\frac{1}{2}$ mark]
$\therefore \Delta AMC \sim \Delta PNR$ (by SAS similarity criterion) [$\frac{1}{2}$ mark]
(ii) $\Delta AMC \sim \Delta PNR$ (from part (i))
$\therefore \angle ACM = \angle PRN$ [$\frac{1}{2}$ mark]
Also $\angle ACB = \angle PRQ$ (as $\Delta ABC \sim \Delta PQR$)
$\therefore \angle ACB - \angle ACM = \angle PRQ - \angle PRN$
$\Rightarrow \angle BCM = \angle QRN$ [$\frac{1}{2}$ mark]
(iii) $\Delta ABC \sim \Delta PQR \Rightarrow \frac{AB}{PQ} = \frac{BC}{QR}$ [$\frac{1}{2}$ mark]
$\Rightarrow \frac{BC}{QR} = \frac{\frac{1}{2} AB}{\frac{1}{2} PQ} \Rightarrow \frac{BC}{QR} = \frac{BM}{QN}$ [$\frac{1}{2}$ mark]
Also $\angle B = \angle Q$ [$\frac{1}{2}$ mark]
$\therefore \Delta BMC \sim \Delta QNR$ (by SAS similarity criterion) [$\frac{1}{2}$ mark]
Given: In $\Delta ABC, DE \parallel BC$
To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$
Construction: Draw $DM \perp AC, EN \perp AB$, join $BE$ and $CD$ (Given + To prove + Construction + Figure: $1$ mark)
Proof: $\frac{ar(\Delta ADE)}{ar(\Delta DBE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \dots (i)$ [$1$ mark]
$\frac{ar(\Delta ADE)}{ar(\Delta ECD)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \dots (ii)$ [$1$ mark]
as $\Delta DBE$ and $\Delta DCE$ lie on the same base $DE$ and between same parallels $BC$ and $DE$
$\therefore ar(\Delta DBE) = ar(\Delta ECD)$ or $\frac{ar(\Delta ADE)}{ar(\Delta DBE)} = \frac{ar(\Delta ADE)}{ar(\Delta ECD)} \dots (iii)$ [$\frac{1}{2}$ mark]
From $(i), (ii)$ and $(iii)$, we get $\frac{AD}{DB} = \frac{AE}{EC}$ [$\frac{1}{2}$ mark]
OR
(b) (i) $\Delta ABC \sim \Delta PQR \Rightarrow \frac{AB}{PQ} = \frac{AC}{PR}$ [$\frac{1}{2}$ mark]
$\Rightarrow \frac{AC}{PR} = \frac{\frac{1}{2} AB}{\frac{1}{2} PQ} \Rightarrow \frac{AC}{PR} = \frac{AM}{PN}$ [$\frac{1}{2}$ mark]
Also $\angle A = \angle P$ [$\frac{1}{2}$ mark]
$\therefore \Delta AMC \sim \Delta PNR$ (by SAS similarity criterion) [$\frac{1}{2}$ mark]
(ii) $\Delta AMC \sim \Delta PNR$ (from part (i))
$\therefore \angle ACM = \angle PRN$ [$\frac{1}{2}$ mark]
Also $\angle ACB = \angle PRQ$ (as $\Delta ABC \sim \Delta PQR$)
$\therefore \angle ACB - \angle ACM = \angle PRQ - \angle PRN$
$\Rightarrow \angle BCM = \angle QRN$ [$\frac{1}{2}$ mark]
(iii) $\Delta ABC \sim \Delta PQR \Rightarrow \frac{AB}{PQ} = \frac{BC}{QR}$ [$\frac{1}{2}$ mark]
$\Rightarrow \frac{BC}{QR} = \frac{\frac{1}{2} AB}{\frac{1}{2} PQ} \Rightarrow \frac{BC}{QR} = \frac{BM}{QN}$ [$\frac{1}{2}$ mark]
Also $\angle B = \angle Q$ [$\frac{1}{2}$ mark]
$\therefore \Delta BMC \sim \Delta QNR$ (by SAS similarity criterion) [$\frac{1}{2}$ mark]