Question
If ABCD is a parallelogram, then the angle bisectors of ∠A and ∠D meet at ‘O’. The measure of ∠AOD is 45∘​
Answer: Option B
:
B
Let∠A=x∘
then ∠D=(180−x)∘
(Since the adjacent angles of a parallelogram sum up to 180∘)
∴∠A2=x2=∠DAO
and∠D2=(180−x)2=90−x2=∠ADO
In ΔAOD,
∠AOD+∠ADO+∠DAO=180∘[Angle sum property of a triangle]
∠AOD=180∘−[∠ADO+∠DAO]
∠AOD=180∘−[x2+90−x2]
or∠AOD=180−90=90∘
∴ The measure of ∠AOD is 90∘
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:
B
Let∠A=x∘then ∠D=(180−x)∘
(Since the adjacent angles of a parallelogram sum up to 180∘)
∴∠A2=x2=∠DAO
and∠D2=(180−x)2=90−x2=∠ADO
In ΔAOD,
∠AOD+∠ADO+∠DAO=180∘[Angle sum property of a triangle]
∠AOD=180∘−[∠ADO+∠DAO]
∠AOD=180∘−[x2+90−x2]
or∠AOD=180−90=90∘
∴ The measure of ∠AOD is 90∘
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