# 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

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