question_answer

If AD is the internal angle bisector of \[\Delta ABC\] with AB = 3 cm and AC = 1 cm, then what is BD : BC?

A) 1 : 3

B) 1 : 4

C) 2 : 3

D) 3 : 4

Correct Answer: D

Solution :

[d] In \[\Delta ABC,\] AD is the internal angle bisector of \[\angle A.\] Using property of internal angle bisector \[\frac{BD}{CD}=\frac{AB}{AC}\] \[\Rightarrow \]   \[\frac{CD}{BD}=\frac{AC}{AB}\] \[\Rightarrow \]   \[\frac{CD}{BD}+1=\frac{AC}{AB}+1\] \[\Rightarrow \]   \[\frac{CD+BD}{BD}=\frac{AC+AB}{AB}\] \[\Rightarrow \]   \[\frac{BC}{BD}=\frac{3+1}{3}\]\[\Rightarrow \]\[\frac{BD}{BC}=\frac{3}{4}\] \[BD:BC=3:4\]