Question

ABCD is a square.E and F are respectively the mid points of BC and CD.If R is the mid point of EF prove that ar(∆AER)=(∆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)
Thank you ☺️☺️☺️

Answer 2

Here's the answer

In ABE and ADF

S - AB = AD (SIDE OF A SQUARE )

A- angle ABE =angle ADF (90)

S- BE = DF ( E and F are the mid points of BC&CD)

Hence , ABE congruent  ADF by SAS

AE=AF (cpct)

Now in  AER & AFR

S - AE= AF (proved above )

S- AR=AR (common)

S- ER= RF (R is the mid points of EF)

HENCE,

AER & AFR are congruent

And ar(AER)=ar(AFR)

                                 proved.