 # If a line is drawn parallel to one side of a triangle `
Question:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio.

Solution:

Given that a in which a line parallel to BC is drawn which meet the other two sides at the point D and E, then we have to prove that

$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$

We have the following diagram with some additional construction. In the above figure, we can see that EF is perpendicular to AB. Therefore EF is the height of the .

Now we are finding the area of and area of Area of $\triangle \mathrm{ADE}=\frac{1}{2}($ base $\times$ hight $)$

$=\frac{1}{2} \mathrm{AD} \cdot \mathrm{EF}$..............(1)

Area of $\triangle \mathrm{DBE}=\frac{1}{2} \mathrm{DB} . \mathrm{EF}$............(2)

Now take the ratio of the two equation, we have

$\frac{\text { Area of } \triangle \mathrm{ADE}}{\text { Area of } \triangle \mathrm{DBE}}=\frac{\frac{1}{2} \mathrm{AD} . \mathrm{EF}}{\frac{1}{2} \mathrm{DB} \cdot \mathrm{EF}}$

$=\frac{\mathrm{AD}}{\mathrm{DB}}$............(3)

Similarly, In $\triangle \mathrm{ADE}$ and $\triangle \mathrm{DEC}$, we have

$\frac{\text { Area of } \triangle \mathrm{ADE}}{\text { Area of } \triangle \mathrm{DEC}}=\frac{\frac{1}{2} \mathrm{AE.DG}}{\frac{1}{2} \mathrm{EC.DG}}$

$=\frac{\mathrm{AE}}{\mathrm{EC}}$...............(4)

But, the area of triangle DBE and triangle DEC are same, therefore equation (4) can be written as

$\frac{\text { Area of } \triangle \mathrm{ADE}}{\text { Area of } \triangle \mathrm{DBE}}=\frac{\frac{1}{2} \mathrm{AE} \cdot \mathrm{DG}}{\frac{1}{2} \mathrm{EC} \cdot \mathrm{DG}}$

$=$..............(5)

From equation (3) and equation (5), we get

$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$

Hence, $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$