58
State and Prove "Basic Proportionality Theorem".
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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]
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]