question_answer

If in a \[\Delta ABC,\,\angle B=2\angle C\] and the bisector of \[\angle B\] meets CA in D, then the ratio BD : DC would be equal to

A)  AD: AC                       

B)  AB: AD

C)  AB: AC           

D)  AC: AB

Correct Answer: C

Solution :

As \[\angle B=2\angle C\] \[\Rightarrow \]   \[\angle ABC=\angle BCA\] In \[\Delta ABC\] and \[\Delta ABD,\]                         \[\angle A=\angle A\] (Common)             \[\angle ABD=\angle BCA\]  (Proved) \[\therefore \]      \[\Delta ABC\sim \Delta ABD\] \[\Rightarrow \]   \[\frac{AB}{AD}=\frac{BC}{BD}=\frac{AC}{AB}\] \[\Rightarrow \]   \[BD:BC=AB:AC\]