In the given figure, I is the incentre ofA ABC.
BI when produced meets the circumcircle of
A ABC at D. Given ZBAC = 55° and ZACB
= 65°; calculate :
AA
(1) ZDCA,
sib /
I
(ii) ZDAC,
,
(iii
) ZDCI,
C
B
(iv) ZAIC.
The
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3
Answer:
Join IA,IC and CD.
(i) IB is bisector of ∠ABC
∠ABD=
2
1
∠ABC=
2
1
(180
∘
−65
∘
−55
∘
)=30
∘
∠DCA=∠ABD=30
∘
-----Angles in the same segment
(ii) ∠DAC=∠CBD=30
∘
-----Angles in the same segment
(iii) ∠ACI=
2
∠ACB
=
2
65
∘
=32.5
∘
-----CI is the angular bisector of ∠ACB
∴∠DCI=∠DCA+∠ACI=30
∘
+32.5
∘
=62.5
∘
(iv) ∠IAC=
2
∠BAC
=
2
55
∘
=27.5
∘
-----AI is the angular bisector of ∠BAC
∠AIC+∠IAC+∠ICA=180
∘
∠AIC=180
∘
−∠IAC−∠ICA=180
∘
−27.5
∘
−32.5
∘
=120
∘
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