Question

please answer the attached question.​

Answer 1

Answer:

Step by step explanation:-

given

AD is the median of ΔABC and E is the midpoint of AD

Through D

 draw DG || BF

In ΔADG

 E is the midpoint of AD and EF || DG

By converse of midpoint theorem we have

F is midpoint of AG and AF = FG  ..............1

Similarly, in ΔBCF 

D is the midpoint of BC and DG || BF   

G is midpoint of CF and FG = GC ..............2

From equations 1 and 2

we will get

AF = FG = GC ........3

 AF + FG + GC = AC

AF + AF + AF = AC (from eu 3)

3 AF = AC

AF = (1/3) AC

Or see attachment

Answer 2

Answer:

Step-by-step explanation

✔️✔️ ANSWER ✔️✔️

AD is the median of ΔABC and E is the midpoint of AD Through D,

draw DG || BF In ΔADG,

E is the midpoint of AD and EF || DG

By converse of midpoint theorem

we have F is midpoint of AG and AF = FG → (1)

Similarly,

in ΔBCF D is the midpoint of BC and DG || BF G is midpoint of CF and FG = GC → (2)

From equations (1) and (2),

we get AF = FG = GC → (3)

From the figure we have,

AF + FG + GC = AC AF + AF + AF = AC

[from (3)] 3 AF = AC AF = (1/3) AC

I HOPE THIS INFO HELPS YOU ☺️☺️☺️

#TEJ