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! $AD$ is not perpendicular to $BC$.

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

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

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

Then consider these two triangles as though their bases lay along the line $BDC$. 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 $BDC" alt="BDC" />.

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

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

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. $AD\perp BC$ 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

$
$

Thanks

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

Prove also taht

$
$

Thanks
Using the fact that the ratio $BDC=c:b" alt="BDC=c:b" /> and $BD+DC = BC =a$, we can say that

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

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

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

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