Question

In triangle ABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ parallel to seg BC. If PQ divides triangle ABC into two equal parts having equal areas, find BP / AB.


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Answer 1

Answer:

Answer : √2-1/√2 is the value of BP/AB

BP/AB = √2-1/√2

Answer 2

Since the line PQ divides △ABC into two equal parts,

area(△APQ)=area(△BPQC)

⇒ area(△APQ)=area(△ABC)−area(△APQ)

⇒ 2area(△APQ)=area(△ABC)

∴ area(△ABC) / area(△APQ) =2/1 --------(1)

Now, in △ABC and △APQ,

∠BAC=∠PAQ [ Common angles ]

∠ABC=∠APQ [ Corresponding angles ]

∴ △ABC∼△APQ [ By AA similarity ]

∴ area(△ABC) / area(△APQ) = AB2/AP2

∴ 2/1=AB2/AP2 [ from (1) ]

⇒ AB / AP=√2/1

⇒ AB−BP / AB =1 / √2

⇒ 1− BP / AB = 1 / √2

⇒ BP / AB = 1 - 1/√2

∴ BP / AB = √2 - 1 / √2