Parallelogram abcd is given. draw line ef so that it goes through the vertex a. point e lies on the side bc and point f lies on the extension of dc . prove that △abe∼△fce.

Proved below by AAA similarity theorem.

Step-by-step explanation:

Line EF is drawn through the vertex A of the parallelogram ABCD. E is on the side BC and F lies on the extension of DC.

Please refer to the attached figure.

In this figure,

In triangles Δ AEB and Δ CEF,

∠ AEB = ∠ CEF (vertically opposite angles)

CF is the extension of side DC.

Since DC || AB, DF || AB and FA is a transversal.

Therefore, ∠ BAE = ∠ EFC (alternate interior angles)

Also, DC || AB and CB is a transversal.

Therefore, ∠ ABE = ∠ FCE (alternate interior angles)

Hence, Δ ABE ~ Δ FCE (By AAA similarity criterian)

multiple all that

step-by-step explanation:

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