Abc s a triangle pq is aline segment intersecting AB in P and AC in Q such that PQnis parallel to BC and divides triangle ABC into two parts in area then BPAB is equal to
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Step-by-step explanation:
Solution
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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(△APQ)
area(△ABC)
=
1
2
---- ( 1 )
Now, in △ABC and △APQ,
∠BAC=∠PAQ [ Common angles ]
∠ABC=∠APQ [ Corresponding angles ]
∴ △ABC∼△APQ [ By AA similarity ]
∴
area(△APQ)
area(△ABC)
=
AP
2
AB
2
∴
1
2
=
AP
2
AB
2
[ From ( 1 ) ]
⇒
AP
AB
=
1
2
⇒
AB
AB−BP
=
2
1
⇒ 1−
AB
BP
=
2
1
⇒
AB
BP
=1−
2
1
∴
AB
BP
=
2
2
−1
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