Question

2. In AABC, AD is median through A and E is
the mid point of AD and BE produced meets
AC in F. Then, AF is equal to​

Answer 1

ANSWER

Given BD=DC=

2

1

BC

AE=ED=

2

1

AD

BX is parallel to BF.

Consider traiangle BFC and DKC

∠BFC=∠DkC, ∠FCD=∠KCD

BFC and DKC are similar .(AAA similarity)

&

KC

FC

=

DC

BC

=

2

1

BC

BC

=2

⇒

KC

FC

=2

⇒

KC=

2

1

FC

K is mid point of FC

FK=KC

....(1)

Consider triangles AEF and AKD

∠EAF=∠DAK ∠BFA=∠DKA (size BF∣∣DK)

∴ΔAFE & ΔAKD are similar (AAA similirity )

⇒

AK

AF

=

AD

AE

=

2AE

AE

=

2

1

⇒

AF=

2

AK

⇒F is mid point of AK

⇒

AF=FK

...(2)

From 1 and 2

FK=KC=AF

∴

AC

AF

=

AF+FK+KC

AF

=

3AP

AF

=

3

1

⇒

AF=

3

1

AC