Mathematics
In the adjoining figure, ABCD is a square. E and F are mid-points of sides BC and CD respectively. If R is mid-point of EF, prove that:
area of ∆AER = area of ∆AFR
Theorems on Area
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Answer
In ∆ABE and ∆ADF,
AB = AD (Sides of a square)
∠B = ∠D (Each angle of a square = 90°)
BE = DF (E is mid-point of BC and F is mid-point of DC)
∴ ∆ABE ≅ ∆ADF (SAS axiom)
∴ AE = AF (c.p.c.t.)
Again in ∆AER and ∆AFR
AE = AF (Proved above)
AR = AR (Common)
ER = FR (R is mid-point of EF)
∴ ∆AER ≅ ∆AFR (SSS axiom)
∴ Area of ∆AER = Area of ∆AFR
Hence, proved that area of ∆AER = area of ∆AFR.
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