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! is not perpendicular to .
Suppose the length of is .
Then, using the formula, write down the areas of 's and . (Hint: the area of .)
Then consider these two triangles as though their bases lay along the line . 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 C" alt="BDC" />.
Then use your first two areas, and you're there.
(Incidentally, although you're told that , it doesn't have to be, provided it is bisected by .)
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. only if the triangle is isosceles.