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Question

I really need help.
Prove: In every triangle, the bisector is located between the median & the height.

Answer 1

triangle: a

b c

let bd, be and bf are the height, bisector and median, respectively.

if bc = ab then bd, be, and bf coincide
1. bca>bac
abd>cbd
abd>abe

2. ce/ea=bc/ab ---> ce
bf=fa then abc = cab
both cbd and cab complement bca to 90 degrees and thus are equal.

Answer 2

In an isosceles triangle, the altitude (height), angle bisector, and median are colinear, they are all the same line. So it is not possible for the bisector to be between the height and the median. Thus we have shown a case for which you are to prove is false. That's all that is necessary to say the original ststement is false.