Question

in triangle ∆ABC PQ is line segment intersecting AB at P and AC atQ such that seg PQ ll to BC , if PQ divide s∆ABC into two equal parts having equal areas finf BP / AB​

Answer 1

Answer:

area(△APQ)=area(△BPQC)

⇒  area(△APQ)=area(△ABC)−area(△APQ)

⇒  2area(△APQ)=area(△ABC)

∴   area(△APQ)area(△ABC)=12     ---- ( 1 )

Now, in △ABC and △APQ,

∠BAC=∠PAQ           [ Common angles ]

∠ABC=∠APQ          [ Corresponding angles ]

∴   △ABC∼△APQ           [ By AA similarity ]

∴  area(△APQ)area(△ABC)=AP2AB2

∴  12=AP2AB2           [ From ( 1 ) ]

⇒  APAB=12

⇒  AB=AB−BP