In $\triangle ABC$, let $D$ be a point on $BC$ such that $AD$ bisects $\angle A$. If $AD = 6$, $BD=4$ and $DC=3$, then find $AB$.

Question

Question:

In $\triangle ABC$, let $D$ be a point on $BC$ such that $AD$ bisects $\angle A$. If $AD = 6$, $BD=4$ and $DC=3$, then find $AB$.

I used the Angle Bisector Theorem and got that $\frac {AB} {AC} = \frac {4} {3}$ . I then said $AB = 4x$ and $BC=3x$ by the Angle Bisector Theorem. That's how far I got. How should I proceed with this?

Answer 1

With angle bisector theorem you can easily find out that $\frac{AB}{AC}=\frac{4}{3}$. By this relationship, you can label sides $AB=4x, AC=3x$

Draw perpendicular $DK, DL$ such that points $D, L$ lies on the sides $AB, AC$ respectively.

You can easily find out that $\triangle AKD \cong \triangle DLA$ (A. A. S.)

By this, you can label $KD=DL=h$ and $AK=AL=a$

By the Pythagorean theorem, you can get the following equations,

$h^2+a^2=36$ ------ (1)

$h^2+(3x-a)^2=9$ ------ (2)

$h^2+(4x-a)^2=16$ ------ (3)

Expanding (2), (3) and substituting (1) as required will lead you to $x=2$

As $AB=4x, AB=2*4=8$

Answer 2

Let $E$ is point of line $AD$ such that $\angle AEB=90°$ and $F$ is point of line $AD$ such that $\angle AFC=90°$. Then triangles ABE and ACF are similar, triangles DBE and DCF are similar. Using this facts, one can write $$\frac{AF}{AE}=\frac{CF}{BE}=\frac{CD}{BD}=\frac{3}{4}$$ $$\frac{DF}{DE}=\frac{CD}{BD}=\frac{3}{4}$$

Then $$\frac{EF}{AE}=1-\frac{AF}{AE}=\frac{1}{4}$$

$$\frac{EF}{DE}=1+\frac{DF}{DE}=\frac{7}{4}$$

$$\frac{AE}{DE}=\frac{EF/DE}{EF/AE}=7$$

$$\frac{AD}{DE}=\frac{AE}{DE}-1=6$$

$$DE=AD/6=1$$ $$AE=7\cdot DE=7$$

$$BE^2=BD^2-DE^2=16-1=15$$

$$AB^2=AE^2+BE^2=49+15=64$$

$$AB=8$$