The sides of a triangle are 50,60 and 70. Find the length of the angle bisector from the longest side to its opposite vertex.
I think the answer is about 42.25, but I don't know how to solve it.
You have a triangle with a segment from one vertex to the opposite side; and you know -- or have enough information to determine -- the lengths of all but one of the segments in the figure. In any problem like that, the use of Stewart's Theorem is a possible path to the solution.
Let the triangle be ABC, and let AD be a segment with D on BC. Then Stewart's Theorem says
Note the theorem holds for any such segment AD -- it can be a median, or an angle bisector, or an altitude; but it can also be ANY segment AD with D on BC.
In your problem, AD is the angle bisector. An angle bisector in a triangle divides the opposite side into two parts whose lengths are in the same ratio as the lengths of the two sides that form the angle.
With AB = 50, AC = 60, and BC = 70, the angle bisector AD divides BC with length 70 into two pieces whose lengths are in the ratio 50:60, or 5:6. That makes the lengths of the two segments BD = 350/11 and CD = 420/11.
Then you are ready to plug in the segment lengths into the formula in Stewart's Theorem:
AB=50; AC=60; BC=70; BD=350/11; CD=420/11;
the unknown x is the length of AD
Plugging in those values leads to the answer of 42.25 that you show for the length of AD. I leave it to you to do the calculations.
