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 is similar fce

The proof is explained step-wise below :

Step-by-step explanation :

For better understanding of the solution see the attached figure :

Given : ABCD is a Parallelogram ⇒ AB ║ DC and AD ║ BC

Now, F lies on the extension of DC. So, AB ║ DF

To Prove : ΔABE is similar to ΔFCE

Proof :

Now, in ΔABE and ΔFCE

∠ABE = ∠FCE ( alternate angles are equal )

∠AEB = ∠FEC ( Vertically opposite angles )

So, by using AA postulate of similarity of triangles

ΔABE is similar to ΔFCE

Hence Proved.

  21.3

step-by-step explanation:

multiplying the given equation by c/sin(25°), we get

  c = 9/sin(25°) ≈ 21.2958 ≈ 21.3

we assume "c" is the length of ab, the side opposite angle c in triangle abc.