Question

AD is the median of triangle ABC and M is the midpoint of AD.Also BM produced meets AC at N. Prove that AN = 1/3 AC

Answer 1

Through D, Draw DK||BN.In triangleADK, M is the mid-point of AD and MN||DK.Therefore, N is the mid-point of AK.Hence, AN=NK.............(i)In triangleBCN, D is the mid-point of BC and BN||DK.Therefore, K is the mid-point of NC.Hence, KC=NK.............(ii)From (i) and (ii),AN=NK=KC..............(iii)Now, AC=AN+NK+KCAC=AN+AN+ANAC=3ANtherefore, AN=1/3AC.

Answer 2

Ito prove 1/3AC=AN
given AD is median of trng ABC,M is m.p of AD
cont DRAW a DF II BM Produce to E
proof- In trng BCE
D is m.p of BC,
DF II BE [ by m.p.t]
F is m.p of CE
so, FE= FC - 1
now, trng ADC
M is m.p of AD
ME II DF
[by m.p.t]
E is m.p of AF
so , AE=EF -2
BY 1 and 2
AE = EF= FC
SO .AN=1/3 AC