In the figure D is the midpoint of side BC of triangle ABC. G is the midpoint of seg AD. Seg BG when produced meets side AC at F. Prove that AF=1/3 AC using Basic proportionality theorem
Answers
Answer:
AF = AC/3.
Step-by-step explanation:
See the attached image for the description of problem.
In ∆ABC, D is mid point of BC. (BD = DC)
G is mid point of AD. (DG = GA)
To prove that AF = AC/3.
Draw a line DE parallel to BF.
In ∆CED and ∆CFB,
BF is common line, DE || BF and angles are same.
Hence ∆CED ~ ∆CFB,
So DC/BD = CE/EF
Since DC = BD, CE = EF. -------------------- E1
In ∆AGF and ∆ADE,
GF || DE, all angles are same.
Hence ∆AGF ~ ∆ADE
So AG/DG = AF/EF
Since AG = DG, AF = EF. ---------------------E2.
From equation E1 and E2, AF = FE = EC.
AF/AC = AF / (AF+FE+EC) = 1/3
AF = AC/3.
Hence proved.
Answer:
B.P.T. means Basis proportionality theorem
Step-by-step explanation:
It is a question for std 10 Maharashtra board 1 . similarity (Maths 2)
Hope it would help you.
