Please help sum11(ii)
ABCD is a parallelogram
the bisectors 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
∠ADC + ∠BCD = 180º (sum of adjacent angles of a parallelogram)
⇒∠ADC/2 +∠ BCD/2 = 90º
⇒∠EDC + ∠ECD = 90º
in triangle ECD sum of angles = 180º
⇒∠EDC + ∠ECD + ∠CED = 180º
⇒ ∠CED = 90º
Hence the first condition is proved that in a parallelogram the bisectors of angles intersect at 90º
Similarly taking triangle BCF it can be proven that ∠BFC=90º
Also ∠BFC+∠CFG = 180º (adjacent angles on a line)
⇒∠CFG = 90º
Now since ∠CFG = ∠CED = 90º it means that lines DE and BG are parallel
Hence it is proved that bisectors of opposite angles in a parallelogram are parallel.
Regards