Question

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]​

Answer 1

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.

Answer 2

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.