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Question
Prove that the bisectors of opposite angles of a parallelogram are parallel.
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Solution
Given ABCD is a parallelogram. The bisectors of ∠ADC and ∠BCD meet at E. The bisectors of ∠ABC and ∠BCD meet at F
From the parallelogram ABCD we have
∠ADC + ∠BCD = 180° ...[ sum of adjacent angles of a parallalogram ]
⇒ `("∠ADC")/(2) + ("∠BCD")/(2)` = 90°
⇒ ∠EDC + ∠EDC = 90°
In triangle ECD sum of angles = 180°
⇒ ∠EDC + ∠ECD + ∠CED = 180°
= ∠CED = 90°
Similarly taking triangle BCF it can prove that ∠BFC = 90°
Now since
∠BFC = ∠CED = 90°
Therefore the lines DE and BF are parallel
Hence proved
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