1. ## trigo(4)

A triangle ABC is right angled at A . The angle bisector of angle A meets BC at D . If BC = a , CA =b , and AB=c , prove that

c/b = BD/DC

i just couldn't get the sketch right .

The line that meets BC at D is not perpendicular right ?

2. Hello thereddevils
Originally Posted by thereddevils
A triangle ABC is right angled at A . The angle bisector of angle A meets BC at D . If BC = a , CA =b , and AB=c , prove that

c/b = BD/DC

i just couldn't get the sketch right .

The line that meets BC at D is not perpendicular right ?
Correct! $\displaystyle AD$ is not perpendicular to $\displaystyle BC$.

$\displaystyle \angle BAD = \angle DAC = 45^o$

Suppose the length of $\displaystyle AD$ is $\displaystyle l$.

Then, using the $\displaystyle \tfrac12ab\sin C$ formula, write down the areas of $\displaystyle \triangle$'s $\displaystyle ABD$ and $\displaystyle ACD$. (Hint: the area of $\displaystyle \triangle ABD = \tfrac12cl\sin 45^o$.)

Then consider these two triangles as though their bases lay along the line $\displaystyle BDC$. So they have a common height. So the ratio of their areas is the same as the ratio of the lengths of their bases which is $\displaystyle BDC$.

Then use your first two areas, and you're there.

(Incidentally, although you're told that $\displaystyle \angle A = 90^o$, it doesn't have to be, provided it is bisected by $\displaystyle AD$.)

3. What are you trying to demonstrate is the angle bisector theorem. Look at Angle bisector theorem - Wikipedia, the free encyclopedia , there is a demonstration for the general case, but I hope you can adapt it. $\displaystyle AD\perp BC$ only if the triangle is isosceles.

4. Originally Posted by thereddevils
A triangle ABC is right angled at A . The angle bisector of angle A meets BC at D . If BC = a , CA =b , and AB=c , prove that

c/b = BD/DC

i just couldn't get the sketch right .

The line that meets BC at D is not perpendicular right ?
This is the continuation of this question which i cant figure out .

Prove also taht

$\displaystyle AD=\frac{\sqrt{2}a \sin B}{1+ \tan B}$

Thanks

5. Hello thereddevils
Originally Posted by thereddevils
This is the continuation of this question which i cant figure out .

Prove also taht

$\displaystyle AD=\frac{\sqrt{2}a \sin B}{1+ \tan B}$

Thanks
Using the fact that the ratio $\displaystyle BDC=c:b$ and $\displaystyle BD+DC = BC =a$, we can say that

$\displaystyle BD = \frac{c}{b+c}\times BC=\frac{ca}{b+c}$

Now use the Sine Rule on $\displaystyle \triangle ABD$:

$\displaystyle \frac{AD}{\sin B}= \frac{BD}{\sin 45^o} = \sqrt2BD$, since $\displaystyle \sin45^o = \frac{1}{\sqrt2}$

$\displaystyle \Rightarrow AD = \frac{\sqrt2ca\sin B}{b+c}= \frac{\sqrt2\sin B}{1+\frac{b}{c}}=\frac{\sqrt2\sin B}{1+\tan B}$