In the figure lines A’A’’ and C’C’’ are parallel. AB is the bisector of ∠ CAA’’ and BC is the bisector of ∠ACC’’. Show that ∠ABC is 90°.
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Answer:
Given: A'A" || C'C"
AB is the angle bisector of
∠
C
A
A
"
and CB is the angle bisector of
∠
A
C
C
"
Let
∠
A
C
C
"=
x
∠
A
C
C
"
+
∠
C
A
A
"=
180
∘
(Since co-interior angles are supplementary)
∠
C
A
A
"=
180
∘
−
x
Since BC is the angle bisector of
∠
A
C
C
"
,
∠
B
C
C
"=
x
2
∠
B
C
A
=
x
2
Similary, since AB is the angle bisector of
∠
C
A
A
"
,
∠
C
A
B
=
180
∘
−
x
2
=
90
∘
−
x
2
∠
B
A
A
"=
180
∘
−
x
2
=
90
∘
−
x
2
In
△
A
B
C
,
x
2
+
(
90
∘
−
x
2
)
+
∠
A
B
C
=
180
∘
⇒
∠
A
B
C
=
90
∘
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