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It is known that the quadrilateral ABCD is a parallelogram, the bisector CF of ∠ BCD intersects AB at F, and the bisector DG of ∠ ADC intersects AB at G 1. Verification: AF = GB 2. Please add another condition on the basis of the known condition, so that Δ EGF is an isosceles right triangle
1) A quadrilateral ABCD is a parallelogram,
∴AB∥≒CD
∠ADG=∠CDG=∠AGD
∠BCF=∠DCF=∠CFB
∴AD=AG,BC=BF
∴AG=BF
∴AF=BG
2) E is the intersection of DG and CF
If Δ EGF is an isosceles right triangle
Then the quadrilateral ABCD should be a rectangle
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