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)

step-by-step explanation:

for this pair of functions g(x)=x-2 and f(x)=|x| is (g o f)(a)= |a| -2

step-by-step explanation:

g(x)=x-2

f(x)=|x|

(g o f)(a)=?

(g o f)(x)=g(f(x))

(g o f)(x)=g(|x|)

(g o f)(x)=|x|-2

(g o f)(a)=|a|-2