In the given figure abcd is a rectangle prove that triangle abe is congruent to triangle cdf
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Δ ABE ≅ Δ CDF In the given figure abcd is a rectangle
Step-by-step explanation:
ABCD is a rectangle
hence AB = CD
∠DAB = 90° & ∠ADC = 90°
in ΔOAD
∠DAE = ∠ODA + ∠AOD ( Exterior angle = Sum of opposite two angles)
=> ∠DAB + ∠BAE = ∠ODA + ∠AOD
=> 90° + ∠BAE = ∠ODA + 90°
=> ∠BAE = ∠ODA
now in Δ EAB & ΔODA
∠BEA = ∠AOD = 90°
∠BAE = ∠ODA
=> Δ EAB ≈ ΔODA
Similarly we can show that
Δ FCD ≈ ΔODA
Δ EAB ≈ ΔODA & Δ FCD ≈ ΔODA
=> Δ EAB ≈ Δ FCD
AB = CD
hence Δ EAB ≅ Δ FCD
=> Δ ABE ≅ Δ CDF
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