At an isosceles triangle two parties are equal, corners at its basis will be equal too. Therefore the bisectors which are carried out to sides will be equal each other. The bisector which is carried out to the basis of an isosceles triangle will be at the same time a median and height of this triangle.
Instruction
1. Let the bisector of AE be carried out to the basis of BC of an isosceles triangle of ABC. The triangle of AEB will be rectangular as the bisector of AE will be at the same time its height. The side of AB will be a hypotenuse of this triangle, and BE and AE - its legs. On Pythagorean theorem (AB^2) = (BE^2)+ (AE^2). Then (BE^2) = sqrt ((AB^2) (-AE^2)). As AE and median of a triangle of ABC, BE = BC/2. Therefore, (BE^2) = sqrt ((AB^2) (-(BC^2)/4)). If the corner at foundation of ABC is set, then from a rectangular triangle the bisector of AE is equal to AE = to AB/sin (ABC). BAE corner = BAC/2 as AE is a bisector. From here, AE = AB/cos(BAC/2).
2. Let BK height to AC side be carried out now. This height is not either a median, or a triangle bisector any more. For calculation of its length exists it is equal to a half of the sum of lengths of all its parties: P = (AB+BC+AC)/2 = (a+b+c)/2, where BC = a, AC = b, AB = with. Stewart's formula for length of the bisector which is carried out to the party of c (that is, AB), will have an appearance: l = sqrt (4abp (p-c)) / (a+b).
3. From Stewart's formula it is visible that the bisector which is carried out to the party of b (AC) will have the same length as b = with.