Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.
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Given: In $\triangle\text{ABC},$ $\text{DE}\parallel\text{BC}$To prove: $\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
Construction: Draw $\text{EF}\perp\text{AB}$ and $\text{DG}\perp\text{AC}$ joine BC and BE.Proof: $\frac{\text{ar}(\triangle\text{ADE})}{{\text{ar}(\triangle\text{DBE})}}=\frac{\frac{1}{2}\text{AD}\times\text{EF}}{\frac{1}{2}\text{DB}\times\text{EF}}=\frac{\text{AD}}{\text{DB}}\ ....(\text{i})$
and $\frac{\text{ar}(\triangle\text{ADE})}{{\text{ar}(\triangle\text{ECD})}}=\frac{\frac{1}{2}\text{AE}\times\text{DG}}{\frac{1}{2}\text{EC}\times\text{DG}}=\frac{\text{AE}}{\text{EC}}\ .....(\text{ii})$ Note that $\triangle\text{DBE}$ and $\triangle\text{ECD}$ are on same base DE and batween same parallel lines DE and BC. $\therefore\text{ar}(\triangle\text{DBE})=\text{ar}(\triangle\text{ECD})\ .......(\text{iii})$ From equations (ii) and (iii), we have $\frac{\text{ar}(\triangle\text{ADE})}{{\text{ar}(\triangle\text{DBE})}}=\frac{\text{AE}}{\text{EC}}\ ........(\text{iv})$ From equtions (i) and (iv), we have $\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$ Hence, proved.
Construction: Draw $\text{EF}\perp\text{AB}$ and $\text{DG}\perp\text{AC}$ joine BC and BE.Proof: $\frac{\text{ar}(\triangle\text{ADE})}{{\text{ar}(\triangle\text{DBE})}}=\frac{\frac{1}{2}\text{AD}\times\text{EF}}{\frac{1}{2}\text{DB}\times\text{EF}}=\frac{\text{AD}}{\text{DB}}\ ....(\text{i})$
and $\frac{\text{ar}(\triangle\text{ADE})}{{\text{ar}(\triangle\text{ECD})}}=\frac{\frac{1}{2}\text{AE}\times\text{DG}}{\frac{1}{2}\text{EC}\times\text{DG}}=\frac{\text{AE}}{\text{EC}}\ .....(\text{ii})$ Note that $\triangle\text{DBE}$ and $\triangle\text{ECD}$ are on same base DE and batween same parallel lines DE and BC. $\therefore\text{ar}(\triangle\text{DBE})=\text{ar}(\triangle\text{ECD})\ .......(\text{iii})$ From equations (ii) and (iii), we have $\frac{\text{ar}(\triangle\text{ADE})}{{\text{ar}(\triangle\text{DBE})}}=\frac{\text{AE}}{\text{EC}}\ ........(\text{iv})$ From equtions (i) and (iv), we have $\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$ Hence, proved.
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