question_answer

In a \[\Delta ABC,\]\[\angle A=80{}^\circ .\] If BD and CD are internal bisectors of \[\angle B\] and \[\angle C\] respectively, then \[\angle BDC\]is equal to

A) \[100{}^\circ \]            

B) \[110{}^\circ \]

C) \[120{}^\circ \]            

D) \[130{}^\circ \]

Correct Answer: D

Solution :

Given, \[\angle BAC=80{}^\circ \]and BD and CD are the bisector of \[\angle B\]and \[\angle C,\]

In \[\Delta ABC,\]

\[\angle BAC+\angle B+\angle C=180{}^\circ \]

[by angle sum property]

\[\Rightarrow \]\[80{}^\circ +\angle B+\angle C=180{}^\circ \]

\[\Rightarrow \]   \[\angle B+\angle C=180{}^\circ -80{}^\circ \]

\[\angle B+\angle C=100{}^\circ \]

Divide by 2 on both the sides, we get

\[\frac{\angle B}{2}+\frac{\angle C}{2}=50{}^\circ \]

i.e.        \[\angle DBC+\angle DCB=50{}^\circ \]

\[\left[ \because \frac{\angle B}{2}=\angle DBC\,\,\text{and}\,\,\frac{\angle C}{2}=\angle DCB \right]\]

Now, in \[\Delta BCO,\]

\[\angle BDC+\angle DBC+\angle DCB=180{}^\circ \]

[by angle sum property form Eq. (i)]

\[\Rightarrow \]   \[\angle BDC+50{}^\circ =180{}^\circ \]

\[\Rightarrow \]\[\angle BDC=180{}^\circ -50{}^\circ =130{}^\circ \]

Alternate Method

In \[\Delta ABC,\]

and BD and CD are internal bisectors of \[\angle B\]and \[\angle C.\]

\[\therefore \]      \[\angle BDC=90{}^\circ +\frac{1}{2}\angle A\] [by results]

\[=90{}^\circ +\frac{1}{2}\times 80{}^\circ =130{}^\circ \]