peaks of the triangle A (3, -5), B (1, -3), C (2, -2). determine the length of its external angle bisectors of the top of the B
You need to remember that the bisector of an exterior angle of a triangle divides externally the opposite side into segments that form a ratio equal to the ratio formed by the other lengths of sides of triangle.
Supposing that BD is the angle bisector of the external angle at B,then `(DA)/(DC) = (AB)/(BC).`
Since the problem provides the coordinates of vertices A,B,C, you may evaluate the lengths AB and BC such that:
`AB = sqrt((x_B - x_A)^2 + (y_B - y_A)^2)`
`AB = sqrt((1 - 3)^2 + (-3 + 5)^2)`
`AB = sqrt(4 + 4) => AB = 2sqrt2`
`BC = sqrt((x_C - x_B)^2 + (y_C - y_B)^2)`
`BC = sqrt((2 - 1)^2 + (-2 + 3)^2)`
`BC = sqrt2`
`(DA)/(DC) = (2sqrt2)/(sqrt2) => (DA)/(DC) = 2`
Hence, evaluating the ratio of the lengths of segments created by the external angle bisector yields (DA)/(DC) = 2.