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Division of a Line Segment

In ABC , PQ i...

Question

$In △ ABC, PQ is a line segment intersecting AB at P and AC at Q such that seg PQ ∥ seg BC . If PQ divides △ ABC into two equal parts means equal in area . Find \frac{BP}{AB} .$

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Solution

$First, we will draw △ ABC in which PQ is a line segment intersecting AB at P and AC at Q, such that PQ ∥ BC and PQ divides △ ABC into two equal parts .$

$Since the line PQ divides △ ABC into two equal parts, A (△ APQ) = A (BPQC) \Rightarrow A (△ APQ) = A (△ ABC) - A (△ APQ) \Rightarrow 2 A (△ APQ) = A (△ ABC) \Rightarrow \frac{A (△ ABC)}{A (△ APQ)} = \frac{2}{1} . . . (1) Now, in △ ABC and △ APQ, ∠ BAC = ∠ PAQ [C ommon angles] ∠ ABC = ∠ APQ [C orresponding angles] Therefore, by the AA similarity test, △ ABC ~ △ APQ Thus, \frac{A (△ ABC)}{A (△ APQ)} = \frac{{AB}^{2}}{{AP}^{2}} = \frac{{BC}^{2}}{{PQ}^{2}} = \frac{{AC}^{2}}{{AQ}^{2}} \Rightarrow \frac{A (△ ABC)}{A (△ APQ)} = \frac{{AB}^{2}}{{AP}^{2}} \Rightarrow \frac{2}{1} = \frac{{AB}^{2}}{{AP}^{2}} [From equation (1)] On taking the square root of both sides, we get : \frac{AB}{AP} = \frac{\sqrt{2}}{1} \Rightarrow \frac{AP}{AB} = \frac{1}{\sqrt{2}} \Rightarrow \frac{AB - BP}{AB} = \frac{1}{\sqrt{2}} \Rightarrow 1 - \frac{BP}{AB} = \frac{1}{\sqrt{2}} \Rightarrow \frac{BP}{AB} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}$

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