In a parallelogram abcd point k belongs to diagonal bd so that bk: dk=1: 4. if the extension of ak meets bc at point e, what is the ratio of be: ec?

BE:EC= 1/3

Step-by-step explanation:

Given ABCD is a parallelogram, Point K is such that it belongs to diagonal BD so that BK:DK=1:4.

If we make an extension of AK which meets BS at point E, then using ΔDKA and ΔEKB, we have

∠DKA=∠EKB (Vertically opposite angles)

∠KDA=∠KBE (Alternate interior angles)

∠DAK=∠BEK (Alternate interior angles)

Thus by AAA similarity,ΔDKA≅ΔEKB

⇒= ,

Since, AD= BC,therefore

= =

Now, BC= BE+EC, ⇒=

⇒1+

⇒

Reciprocating on both the sides, we get

for the first question it would be the second answer for the second question it would be the third answer and for the third question it would be the first answer

step-by-step explanation:

you should at least use a calculater

step-by-step explanation: