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
Solution:
Given, a line is drawn parallel to one side of a triangle to intersect the other two sides.
We have to prove that the two sides are divided in the same ratio.
Consider a triangle ABC, D and E are the points that intersect AB and AC.
DE is parallel to BC.
Now join BE and CD.
Also, draw EF and DG perpendicular to AB and AC.
We know that, area of triangle = 1/2 × base × height
Now, area of △ADE = 1/2 × AD × EF
Area of △BDE = 1/2 × BD × EF
So, area of △ADE/area of △BDE = (1/2 × AD × EF)/(1/2 × BD × EF) = AD/BD ------------ (1)
Area of △ADE = 1/2 × AE × DG
Area of △DEC = 1/2 × EC × DG
So, area of △ADE/area of △DEC = (1/2 × AE × DG)/(1/2 × EC × DG) = AE/EC ----------- (2)
From the figure, △BDE and △DEC lie on the same base DE.
Also, △BDE and △DEC lie between the same parallel DE and BC.
So, area of △BDE = area of △DEC
Since area of the triangles are same, then
AD/BD = AE/EC
Therefore, the two sides are divided in the same ratio.
✦ Try This: Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side of BC. If AC and BD intersect at P, prove that AP × PC = BP × PD.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 6
NCERT Exemplar Class 10 Maths Exercise 6.4 Problem 3
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
Summary:
It is proven 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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