5. In the figure, D is the midpoint of side
BC of AABC. G is the midpoint of seg
AD. seg BG when produced meets
side AC at F. Prove AF = AC using
basic proportionality theorem. (HOTS)
Hint : Draw line parallel to BF passing through D]
Answers
Answer:
Step-by-step explanation:
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.
Proved , AF = 3AC.
• Drawn the diagram and constructed an line DH || BF .
• consider the theorm,
If a line is drawn parallel to one
side of a triangle to intersect the
other two sides in distinct points,
the other two sides are divided in
the same ratio.
• In triangle CDH and triangle CBF
DH || BF & BD = DC (D is mid pt.)
• So, CH= FH ( By theorm)___(1)
• In triangle ADH and triangle AGF
DH || GF & GD = AG (G is mid pt.)
• So, AF= FH ( By theorm)___(2)
• From (1) &(2)
• AF = FH = CH
• Also, AC = AF + FH + CH
• AC = 3AF
Hence proved.
