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  • ABCD is a square. E and F are respectively the midpoints of BC and CD. If R is the mid-point of EF (Fig.), prove that ar (AER) = ar(AFR)

ABCD is a square. E and F are respectively the midpoints of BC and CD. If R is the mid-point of EF (Fig.), prove that ar (AER) = ar(AFR)

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Solution.

Given that ABCD is a square
E and F are the midpoints of BC and CD
R is the mid-point of EF
Proof: Consider \triangle ABE and \triangle ADF
We know that BC = DC              (All sides of square are equal)
\Rightarrow \frac{BC}{2}=\frac{DC}{2}
\Rightarrow EB=DF                         (Since E and F are mid points of BC and CD)
\angle ABE=\angle ADF=90^{\circ}     (All angles in a square are right angle)
\Rightarrow AB=AD                        (All sides in a square are equal)
\therefore \triangle ABE\cong \triangle ADF         (By SAS criterion)
Hence, AE = AF (by C.P.C.T)
Now, ER = RF ( R is the midpoint of EF)
AR = AR (Common)
\therefore \triangle AER\cong \triangle AFR (By SSS criterion)
Hence, AR divides the triangle into two triangles of equal area.
\Rightarrow ar(\triangle AER)=ar(\triangle AFR)
Hence proved.

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