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  • 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.

Q: 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.

Answers (1)

Let ABC is a triangle in which line DE is parallel to BC which interests lines AB and AC at D and E.

To prove:- line DE divides both sides in the same ratio.

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

Construction:-

\text{Join BE , CD and draw } EF\perp AB and DG\perp AC

Proof:-

\frac{ar(\Delta ADE)}{ar(\Delta BDE)}=\frac{\frac{1}{2}\times AD \times EF}{\frac{1}{2}\times DB \times EF}

\frac{ar(\Delta ADE)}{ar(\Delta BDE)}=\frac{AD}{DB}\; \; \; \; \; \; \; \; .....(1)

Similarly

\frac{ar(\Delta ADE)}{ar(\Delta DEC)}=\frac{\frac{1}{2}\times AE \times GD}{\frac{1}{2}\times EC \times GD}=\frac{AE}{EC}\; \; \; \; \; \; \; \; ...(2)

Also

ar\left ( \Delta BDE \right )=ar\left ( \Delta DEC \right )\; \; \; \; \; ....(3)

[\Delta BDE\ and \ \Delta DEC\ \text{lie between the same parallel lines DE and BC and on the same base DE ]} 

from equation (1), (2) and (3)

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

Hence proved

Posted by

infoexpert23

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