The bisectors of angles A and B of parallelogram ABCD meet at point F of side CD. Prove that F is the midpoint of CD.
| August 7, 2021 | education
AF is the bisector of ∟А, then ∟КAF = ∟ВAF, ∟ВAF = ∟AFD – as internal criss-crossing at parallel AB and CD and secant AF, so ∟КAF = ∟AFD. ∆ AFD – isosceles, AD = FD. The situation is similar with ∆ ВFC, ВС = FС. ABCD is a parallelogram, therefore AD = BC, therefore FD = FС, F is the middle of CD.
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