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State true or false.

If a line passing through two sides of triangle and it is parallel to third side then divides other two sides in same ratio.

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True

False

answer is A.

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Detailed Solution

Let's assume an ADE triangle.
In

△ ADE

(Baes AD)

Triangle’s area

= \frac{1}{2} \cdot AD \cdot ME

…….. (1)
Again in

△ ADE

(Base AE)

ar \cdot (△ ADE) = \frac{1}{2} \cdot AE \cdot ND

…….. (2)

ar \cdot (△ BDE) = \frac{1}{2} \cdot DB \cdot ME

…….. (3)
In

△ DEC (Base EC)

ar \cdot (△ DEC) = \frac{1}{2} \cdot EC \cdot ND

…….. (4)
Adding to Equation (1) & (3)

\frac{ar \cdot ΔADE}{ar \cdot ΔBDE} = \frac{\frac{1}{2} \cdot AD \cdot ME}{\frac{1}{2} \cdot DB \cdot ME}

\frac{ar △ ADE}{ar \cdot △ BDE} = \frac{AD}{DB}

…….. (5)
Adding to equation (2) & (4)

\frac{aar (△ ADE)}{ar \cdot (△ DEC)} = \frac{\frac{1}{2} \cdot AE \cdot ND}{\frac{1}{2} EC \cdot ND}

\frac{ar (△ ADE)}{△ DEC} = \frac{AE}{EC}

…….. (6)
By theorem,

ar \cdot (△ BDE) = ar \cdot (△ DEC) \frac{ar (△ ADE)}{ar \cdot (△ BDC)} = \frac{ar \cdot (△ ADE)}{ar \cdot (△ DEC)} = \frac{AD}{BD}

\frac{ar (△ ADE)}{ar (△ DEC)} = \frac{AD}{BD}

…….. (7)
By equation (6) & (7)
Here, left hand side is equal to the right-hand side.

\frac{AE}{EC} = \frac{AD}{BD}

Thus,

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

Hence, the given statement is true.
Option 1 is correct.

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