Question

In $\Delta ABC$, if AD is the bisector of $\angle A$, prove that
$\dfrac{{Area\left( {\Delta ABD} \right)}}{{Area\left( {\Delta ACD} \right)}} = \dfrac{{AB}}{{AC}}$

Answer 1

Hint: Start by using the information that is given to us that is AD is the bisector of $\angle A$, using this we can obtain a relation. Then we draw a perpendicular and find the ratio asked in the question.

Complete step-by-step answer:

In $\Delta ABC$, AD is the bisector of $\angle A$, by internal angle bisector theorem, the bisector of vertical angle of a triangle divides the base in the ratio of the other two sides, therefore,
$\dfrac{{AB}}{{AC}} = \dfrac{{BD}}{{DC}}$ ….. (i)
From A, draw $AL \bot BC$.

$ \Rightarrow \dfrac{{Area\left( {\Delta ABD} \right)}}{{Area\left( {\Delta ACD} \right)}} = \dfrac{{\dfrac{1}{2}BD \cdot AL}}{{\dfrac{1}{2}DC \cdot AL}}\left[ {\because Area = \dfrac{1}{2} \times b \times h} \right]$
Cancel out the common terms,
$ \Rightarrow \dfrac{{Area\left( {\Delta ABD} \right)}}{{Area\left( {\Delta ACD} \right)}} = \dfrac{{BD}}{{DC}}$
From (i), we can say that DC=AC, therefore,
$ \Rightarrow \dfrac{{Area\left( {\Delta ABD} \right)}}{{Area\left( {\Delta ACD} \right)}} = \dfrac{{BD}}{{AC}}$ …..(From (i))
Hence Proved!

Note: In geometry, the angle bisector theorem is concerned with the relative lengths of the two segments that a triangle's side is divided into by a line that bisects the opposite angle. It equates their relative lengths to the relative lengths of the other two sides of the triangle.