Respuesta :
Answer:
Given: An Isosceles Triangle ABC with a vertex at B.
Midpoint M of the base AC.
To Prove: BM is perpendicular to AC.
Proof:
Let the coordinates of the points of the isosceles triangle be given as:
A = (-k, 0)
Vertex, B = (0,a)
C = (k, 0)
Midpoint, M = (0,0)
Slope of the base segment, AC:
[tex]=\dfrac{dy}{dx} = \dfrac{0 - 0}{k - (-k)} = \dfrac{0}{2\cdot k}[/tex]
Slope of the base segment, AC= [tex]\dfrac{0}{2\cdot k}=0[/tex]
Slope of the segment that joins the vertex angle of an isosceles triangle to the midpoint of its base, BM.
[tex]\text{Slope of BM =} \dfrac{0 - a}{0 - 0} = \dfrac{-a}{0}\\[/tex]
[tex]\text{Slope of BM = } \dfrac{-a}{0}[/tex] = Undefined
Two lines are perpendicular if the gradient of one is a negative reciprocal of the other.
Since [tex]-\dfrac{a}{0}[/tex] is a negative reciprocal of 0 for arbitrary values of a, BM and AC are perpendicular.
This concludes the proof.
