Question

In the given figure, $I$ is the incentre of $\Delta ABC$. $AI$ produced meets the circumcircle of $\Delta ABC$ at $D$; $\angle ABC={55}^o$ and $\angle ACB={65}^o$. Then (i) $\angle BCD$ (ii) $\angle CBD$ (iii) $\angle DCI$ (iv) $\angle BIC$ are respectively:

• A. ${50}^o,{20}^o,{32.5}^o\&{240}^o$
• B. ${30}^o,{30}^o,{62.5}^o\&{120}^o$
• C. ${35}^o,{35}^o,{82.5}^o\&{100}^o$
• D. ${40}^o,{20}^o,{32.5}^o\&{150}^o$

Answer 1

Answer: (B)

${30}^o,{30}^o,{62.5}^o\&{120}^o$

$Given-The\Delta ABC,whoseincentreisI,hasbeeninscribedinacicle.AIisproducedtomeetthecircumferenceatD.BD\&DChavebeenjoined.\angle ABC={55}^o\&\angle ACB={65}^o.Tofindout-\left(i\right)\angle BCD=?\left(ii\right)\angle CBD=?\left(iii\right)\angle DCI=?\left(iv\right)\angle BIC=?Solution-\angle ABC=\angle ADC={55}^o\left[sinceboththeangleshavebeensubtendedbythechordACtothecircumferenceatB\&D\right].Also,\angle ACB=\angle ADB={65}^o\left[sinceboththeangleshavebeensubtendedbythechordABtothecircumferenceatC\&D\right].So,\angle BDC=\angle ADC+\angle ADB={55}^o+{65}^o={120}^o.Now,reflex\angle BIC=2\angle BDC\left(sincereflex\angle BIC\&\angle BDCareanglesatthecentre\&angleatthecircumferenceatDsubtendedbythechordBC\right).\therefore reflex\angle BIC=2\times {120}^o={240}^o⟹\angle BIC={360}^o-{240}^o={120}^o.Bythesamereasoning\angle BAC=\frac{1}{2}\times \angle BIC=\frac{1}{2}\times {120}^o={60}^o.Again,Iistheincentreof\Delta ABC.\therefore AI,BI\&CIareangularbisectorof\angle BAC,\angle ABC\&\angle ACBrespectively.\therefore \angle BAI=\frac{1}{2}\times \angle BAC=\frac{1}{2}\times {60}^o={30}^o=\angle CAI.So,\angle BCD=\angle BAI={30}^o\left[since\angle BCD\&\angle BAIareanglesatthecentre\&angleatthecircumferenceatAsubtendedbythechordBD\right].Bythesamereasoning\angle CBD=\angle CAI={30}^o.Again,CIistheangularbisectorof\angle ACD.\therefore \angle BCI=\angle ACI=\frac{1}{2}\times {65}^o={32.5}^o.\therefore \angle DCI=\angle BCD+\angle BCI={32.5}^o+{30}^o={62.5}^o.So\angle BCD,\angle CBD,\angle DCI\&\angle BICare{30}^o,{30}^o,{62.5}^o\&{120}^{o}respectively.Hence,optionBiscorrect.$