Question

ABCD is a parallelogram with AD produced to its own lenght to E.BE and DE intersect at K. AK is produced toCE meets at H.Prove that KH=1/3AK

Answer 1

Answer:

Refer to the attached figure.

Since, ABCD is a parallelogram, therefore AB is parallel to CD.

Therefore, AF is parallel to CD.

And we can observe that EF is a traversal.

\angle 2 = \angle 4 (Corresponding angles) (Equation 1)

Therefore, AE is parallel to BC.

So, \angle 1 = \angle 3 (Corresponding angles) (Equation 2)

Now, in triangle DEC,

DE = DC

Therefore,

So, \angle 1 = \angle 2 (Equation 3)

(Angles opposite to the equal opposite sides are also equal)

Therefore, \angle 3 = \angle 4 (By equations 1,2 and 3)

Therefore, BC = BF

Hence, proved.

hope it will help you

Answer 2

Answer:

sorry for irrevelant answer mark as brainliest to my siso....