56
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
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Correct figure [$\frac{1}{2}$ mark]
Given: In $\Delta ABC, DE \parallel BC$
To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$
Construction: Join BE, DC, Draw $DM \perp AC$ and $EN \perp AB$ [$1$ mark]
Proof : $\frac{ar(\Delta ADE)}{ar(\Delta BDE)} = \frac{\frac{1}{2} \times AD \times EF}{\frac{1}{2} \times DB \times EF} \Rightarrow \frac{ar(\Delta ADE)}{ar(\Delta BDE)} = \frac{AD}{DB}$ ..........(i) [$1$ mark]
and $\frac{ar(\Delta ADE)}{ar(\Delta CDE)} = \frac{\frac{1}{2} \times AE \times DG}{\frac{1}{2} \times EC \times DG} \Rightarrow \frac{ar(\Delta ADE)}{ar(\Delta CDE)} = \frac{AE}{EC}$ ................... (ii) [$1$ mark]
As $\Delta BDE$ and $\Delta CDE$ are on the same base DE and between the same parallels DE and BC.
$\therefore ar(\Delta BDE) = ar(\Delta CDE)$ ..……………..(iii) [$1$ 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: Join BE, DC, Draw $DM \perp AC$ and $EN \perp AB$ [$1$ mark]
Proof : $\frac{ar(\Delta ADE)}{ar(\Delta BDE)} = \frac{\frac{1}{2} \times AD \times EF}{\frac{1}{2} \times DB \times EF} \Rightarrow \frac{ar(\Delta ADE)}{ar(\Delta BDE)} = \frac{AD}{DB}$ ..........(i) [$1$ mark]
and $\frac{ar(\Delta ADE)}{ar(\Delta CDE)} = \frac{\frac{1}{2} \times AE \times DG}{\frac{1}{2} \times EC \times DG} \Rightarrow \frac{ar(\Delta ADE)}{ar(\Delta CDE)} = \frac{AE}{EC}$ ................... (ii) [$1$ mark]
As $\Delta BDE$ and $\Delta CDE$ are on the same base DE and between the same parallels DE and BC.
$\therefore ar(\Delta BDE) = ar(\Delta CDE)$ ..……………..(iii) [$1$ mark]
From (i), (ii) and (iii), we get $\frac{AD}{DB} = \frac{AE}{EC}$ [$\frac{1}{2}$ mark]