Prove that the bisectors of opposite angles of a parallelogram are parallel.
ABCD is a parallelogram, the bisector of ∠ADC and ∠BCD meet at point E and the bisectors of ∠BCD and ∠ABC meet at F. We have to prove that the ∠CED=90o and ∠CFG=90o. This way we will be able to prove that DE and CE intersect at right angles and BG and ED are parallel. This way we will be able to prove that DE and CE intersect at right angles and BG and ED are parallel.
∠ADC+∠BCD=180o [sum of adjacent angles of a parallelogram]
⇒∠ADC2+∠BCD2=90o
∴∠EDC+∠ECD=90o
In △ECD sum of angle =180o
⇒∠EDC+∠ECD+∠CED=180o
∴∠CED=90o
Hence, the first condition is proved that in a parallelogram the bisectors of angles intersect at 90o
In △ECD, sum of angles =180o
⇒∠EDC+∠ECD+∠CED=180o
∠CED=90o
Hence, the first condition is proved that in a parallelogram, the bisectors of angle intersect at 90o
Similarly taking ABCF,∠BFC=90o
∠BFC+∠CFG=180o
∠CFG=90o
∴DE||BG
Hence proved