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(A) 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.
OR
(B) In a $\Delta ABC$, P and Q are points on AB and AC respectively such that $PQ \parallel BC$. Prove that the median AD, drawn from A to BC, bisects PQ.
OR
(B) In a $\Delta ABC$, P and Q are points on AB and AC respectively such that $PQ \parallel BC$. Prove that the median AD, drawn from A to BC, bisects PQ.
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(A) 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]
OR
(B) Correct figure [$\frac{1}{2}$ mark]
Given $PQ \parallel BC \Rightarrow PR \parallel BD$
Therefore in $\Delta APR$ and $\Delta ABD$, $\angle PAR = \angle BAD$ and $\angle APR = \angle ABD$ [$1$ mark]
$\therefore \Delta APR \sim \Delta ADB$ (By AA similarity criterion)
$\Rightarrow \frac{AR}{AD} = \frac{PR}{BD}$ (i) [$\frac{1}{2}$ mark]
Similarly $\Delta ARQ \sim \Delta ADC \Rightarrow \frac{AR}{AD} = \frac{RQ}{DC}$ (ii) [$1$ mark]
Using (i) and (ii) $\frac{PR}{BD} = \frac{RQ}{DC}$ [$\frac{1}{2}$ mark]
AD is the median $\therefore BD = DC \Rightarrow PR = RQ$ [$1$ mark]
i.e. AD bisects PQ [$\frac{1}{2}$ 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]
OR
(B) Correct figure [$\frac{1}{2}$ mark]
Given $PQ \parallel BC \Rightarrow PR \parallel BD$
Therefore in $\Delta APR$ and $\Delta ABD$, $\angle PAR = \angle BAD$ and $\angle APR = \angle ABD$ [$1$ mark]
$\therefore \Delta APR \sim \Delta ADB$ (By AA similarity criterion)
$\Rightarrow \frac{AR}{AD} = \frac{PR}{BD}$ (i) [$\frac{1}{2}$ mark]
Similarly $\Delta ARQ \sim \Delta ADC \Rightarrow \frac{AR}{AD} = \frac{RQ}{DC}$ (ii) [$1$ mark]
Using (i) and (ii) $\frac{PR}{BD} = \frac{RQ}{DC}$ [$\frac{1}{2}$ mark]
AD is the median $\therefore BD = DC \Rightarrow PR = RQ$ [$1$ mark]
i.e. AD bisects PQ [$\frac{1}{2}$ mark]