Instruction

1

Let the bisector of AE is drawn to the base BC of isosceles

**triangle**ABC. Triangle AEB will be rectangular, since the angle bisector AE will be its height. Side AB is the hypotenuse of this**triangle**, and BE and AE for his legs.By the Pythagorean theorem AB^2) = (YZ^2)+(AE^2). Then (BE^2) = sqrt (AB^2)-(AE^2)). Since AE and median**of the triangle**ABC, BE = BC/2. Therefore, (BE^2) = sqrt (AB^2)-((BC^2)/4)).If you specify the base angle ABC, then from rectangular**triangle**the angle bisector AE equal AE = AB/sin(ABC). Angle BAE = BAC/2, since AE is the bisector. Hence, AE = AB/cos(BAC/2).2

Suppose now that the height of BK is conducted to the side AC. This height is neither a median nor a bisector

**of the triangle**. To calculate its length there is a formula Stewart.The perimeter**of a triangle**is the sum of the lengths of all its sides is P = AB+BC+AC. And properiter equal to half the sum of the lengths of all its sides: P = (AB+BC+AC)/2 = (a+b+c)/2 where BC = a, AC = b, AB = c.Stuart the formula for the length of the bisector drawn to the side c (i.e., AB), will be of the form: l = sqrt(4abp(p-c))/(a+b).3

From the formula of Stuart can be seen that the angle bisector drawn to the side b (AC), will have the same length as b = c.