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  • 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.    

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.

 

 

Answers (1)

Let PQR is a triangle in which XY \parallel QR, XY intersects PQ & PR at X and Y respectively.

Then to prove

 \frac{PX}{XQ} = \frac{PY}{YR}

area of triangle = \frac{1}{2}\times b\times h

or (\Delta PXY) = \frac{1}{2}\times PX\times YN \qquad -(1)

or (\Delta PXY) = \frac{1}{2}\times PY\times XN \qquad -(2)

or (\Delta QXY) = \frac{1}{2}\times QX\times NY \qquad -(3)

or (\Delta RXY) = \frac{1}{2}\times YR\times XM \qquad -(4)

On divinding (1) by (3) we get

        \frac{ar(\Delta PXY)}{ar(\Delta QXY)} = \frac{PX}{QX}\qquad -(5)

Again, divide (2) by (4) we get

      \frac{ar(\Delta PXY)}{ar(\Delta RXY)} = \frac{PY}{YR}\qquad -(6)

Since the area of triangles with the same base and between the same parallel lines are equal. So, {ar(\Delta QXY)} = {ar(\Delta RXY)}

[as {ar(\Delta QXY)}\ \& \ {ar(\Delta RXY)} are on same base XY & between same parallel lines XY & QR]

From equations (5), (6) and (7) we get

        \frac{PX}{XQ} = \frac{PY}{YR}

Hence, proved.

Posted by

Safeer PP

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