Question

D is the mid point of side BC of triangle ABC and E is the mid point of side AD. BE produced meets AC at K. Prove that AK=1/3 AC

Answer 1

it is the answer of the question

Answer 2

Given :
triangle ABC, in which D is the mid point of side BC and E is the mid point of side AD. BE is produced meeting AC at K.

To prove: AK = 1/3 of AC

Construction:
DF || BK is drawn meeting AC at F

Proof:
In triangle ADF,
Since DF || EK,
By B.P theorem, we get,
AE / ED = AK / KF
=> AE / AE = AK / KF [AE = ED]
=> 1 = AK / KF
=> AK = KF --- (1)

Again,
In triangle BCK,
Since BK || DF,
By B.P theorem, we get,
CD / DB = FC / KF
=> CD / CD = FC / KF [CD = DB]
=> 1 = FC / KF
=> FC = KF --- (2)

from (1) and (2), we get,
AK = KF = FC --- (3)

and AK + KF + FC = AC
=> AK + AK + AK = AC [from (3)]
=> 3AK = AC
=> AK = 1/3 of AC
hence proved