# Abcd is a Parallelogram. E is a Point on Ba Such that Be = 2 Ea and F is a Point on Dc Such that Df = 2 Fc. Prove that Ae Cf is a Parallelogram Whose Area is One Third of the Area of Par - Mathematics

ABCD is a parallelogram. E is a point on BA such that BE = 2 EA and F is a point on DC
such that DF = 2 FC. Prove that AE CF is a parallelogram whose area is one third of the
area of parallelogram ABCD.

#### Solution

Construction: Draw FG ⊥ AB
Proof: We have
BE = 2EA and DF = 2 FC

⇒ AB - AE = 2EA and DC - FC = 2FC

⇒ AB  = 3EA  and DC  = 3FC

⇒  AE = 1/2 AB and FC =  1/3 CD   .......... (1)

But AB  = DC
Then, AE = DC    [opposite sides of ||gm]
Then, AE = FC

Thus, AE = FC and AE || FC.
Then, AECF is a parallelogram

Now ar (||gm  = AECF)  =  AE × FG

⇒  ar (||gm  =AECF) =1/3 ABxx FG  form       ....... (1)

⇒  3ar (||gm  =AECF) = AB × FG                   ........(2)

and area  (||gm  =ABCD) = AB  × FG          ........... (3)

Compare equation (2) and (3)

⇒  3 ar  (||gm  =AECF) = area   (||gm  =ABCD)

⇒ area (||gm  =AECF) = 1/3 area (||gm  =ABCD)

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 14 Areas of Parallelograms and Triangles
Exercise 14.3 | Q 18 | Page 46