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 ?
Hello thereddevilsCorrect! $\displaystyle AD$ 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 $\displaystyle AD$.)
Grandad
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.
Hello thereddevilsUsing 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 $\displaystyle \sin45^o = \frac{1}{\sqrt2}$
$\displaystyle \Rightarrow AD = \frac{\sqrt2ca\sin B}{b+c}= \frac{\sqrt2\sin B}{1+\frac{b}{c}}=\frac{\sqrt2\sin B}{1+\tan B}$
Grandad