Question 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.
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Solution
Consider ΔABC in which a line DE parallel to BC intersects AB at D and AC at E. To prove that DE divides the two sides in the same ratio. i.e., ABDB=AEEC
Construction Join BE, CD and draw EF ⊥ AB and DG ⊥ AC. Proof Here, ar(ΔADE)ar(ΔBDE)=12×AD×EF12×DB×EF[∵areaoftriangle=12×base×height] =ADDB ….(i) Similarly, ar(ΔADE)ar(ΔDEC)=12×AE×GD12×EC×GD=AEEC …..(ii) Now, since, ΔBDE and ΔDEC lie between the same parallel DE and BC and on the same base DE. So, ar(ΔBDE)=ar(ΔDEC) ……(iii) From Eq.s (i), (ii) and (iii), we get ADDB=AEEC