in the given figure I is the incenter of triangle abc here Ai produced meets the circumcircle of triangle abc at d.
Answers
Answer:
ANSWER
Given−
TheΔABC,whoseincentreisI,hasbeeninscribedinacicle.
AIisproducedtomeetthecircumferenceatD.
BD&DChavebeenjoined.
∠ABC=55
o
&∠ACB=65
o
.
Tofindout−
(i)∠BCD=?(ii)∠CBD=?(iii)∠DCI=?(iv)∠BIC=?
Solution−
∠ABC=∠ADC=55
o
[sinceboththeangleshavebeensubtended
bythechordACtothecircumferenceatB&D].
Also,∠ACB=∠ADB=65
o
[sinceboththeangleshavebeensubtended
bythechordABtothecircumferenceatC&D].
So,∠BDC=∠ADC+∠ADB=55
o
+65
o
=120
o
.
Now,reflex∠BIC=2∠BDC(sincereflex∠BIC&∠BDCareangles
atthecentre&angleatthecircumferenceatDsubtendedby
thechordBC).
∴reflex∠BIC=2×120
o
=240
o
⟹∠BIC=360
o
−240
o
=120
o
.
Bythesamereasoning∠BAC=
2
1
×∠BIC=
2
1
×120
o
=60
o
.
Again,IistheincentreofΔABC.
∴AI,BI&CIareangularbisectorof
∠BAC,∠ABC&∠ACBrespectively.
∴∠BAI=
2
1
×∠BAC=
2
1
×60
o
=30
o
=∠CAI.
So,∠BCD=∠BAI=30
o
[since∠BCD&∠BAIareangles
atthecentre&angleatthecircumferenceatAsubtendedby
thechordBD].
Bythesamereasoning∠CBD=∠CAI=30
o
.
Again,CIistheangularbisectorof∠ACD.
∴∠BCI=∠ACI=
2
1
×65
o
=32.5
o
.
∴∠DCI=∠BCD+∠BCI=32.5
o
+30
o
=62.5
o
.
So∠BCD,∠CBD,∠DCI&∠BICare30
o
,30
o
,62.5
o
&120
o
respectively.
Hence,optionBiscorrect.
solution
