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)
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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)
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