question_answer

If AD is the internal angular bisector of \[\Delta \mathbf{ABC}\] with \[\mathbf{AB}=\mathbf{4}\]cm and \[\mathbf{AC}=\mathbf{1}\]cm, then what is BD : BC ?

A)  1 : 3                            

B)  1 : 4             

C)  5 : 4                            

D)  4 : 5

Correct Answer: D

Solution :

(d):- In \[\Delta ABC\] AD is the internal angle bisector of \[\angle A\]. Using property of internal angle bisector. \[\frac{BD}{CD}=\frac{AB}{AC}\Rightarrow \frac{CD}{BD}=\frac{AC}{AB}\] \[\Rightarrow \] \[\frac{CD}{BD}\text{+1=}\frac{AC}{AB}\text{+1}\] \[\Rightarrow \frac{CD+BD}{BD}=\frac{AC+AB}{AB}\] \[\Rightarrow \]\[\frac{BC}{BD}=\frac{1+4}{4}\Rightarrow \frac{BD}{BC}=\frac{4}{5}\] \[\therefore \]\[BD:BC=4:5\]