In a parallelogram abcd, ab = 10 cm and ad = 6 cm. the bisector of ∠a meets dc in
e. ae and bc produced meet at f. find the length of cf
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When u see the figure angle bar=angle daf,angle d=angle b so from SS similarity
∆ade ~∆abf
angle f = angle aed
& angle aed = angle cef (bcoz of vertically opposite angles)
So,angle f = angle cef
So,CE =CF
now ∆ade ~∆abf
So, ab/ad=bf/de
10/6=(bc+cf)/(dc-ec)
10(ab-cf)=6(ad+cf)
10(10-cf)=6(6+cf)
100-36=6cf+10cf
64/16=cf
cf=4cm
∆ade ~∆abf
angle f = angle aed
& angle aed = angle cef (bcoz of vertically opposite angles)
So,angle f = angle cef
So,CE =CF
now ∆ade ~∆abf
So, ab/ad=bf/de
10/6=(bc+cf)/(dc-ec)
10(ab-cf)=6(ad+cf)
10(10-cf)=6(6+cf)
100-36=6cf+10cf
64/16=cf
cf=4cm
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