Question

Abcd is a square. E and f respectively the mid points of bc and cd. If r is the mid point of ep. Prove that ar(aer)=ar(afr)

Answer 1

Given:- ABCD is a square. E and F are respectively the midpoints BC and CD. R is the midpoints of EF.

To prove:- ar(Triangle AER) =ar(triangle AFR)

Proof:- In triangle ABE and triangle ADF

AB=AD[sides of a square are equal]

angle ABE = angle ADF [each 90°]

E is the midpoint of BC and F is the

midpoints of CD. [1\2 BC=1/2CD]

by SAS rule,

ar(triangle ABE) is congruent to

ar(triangle ADF)

therfore AE=AF (c. P. C. T) -1

now in triangle AER and triangle AFR

AE=AF[from 1]

ER=RF(R is the midpoint of ED)

AR=AR(common side)

by SSS rule

Triangle AER congruent to triangle

AFR.

Hence( triangle AER) = ( Triangle

AFR)