SOLUTION: prove that the line drawn from the vertex of an isosceles triangle through the point of intersection of 2 medians drawn from the base angle is perpendicular to the base. Explain wi

Question 1208950: prove that the line drawn from the vertex of an isosceles triangle through the point of intersection of 2 medians drawn from the base angle is perpendicular to the base. Explain with statement s and reason
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
Triangle ABC is an isosceles triangle, so sides AC and BC are congruent, and congruent, and so are triangle ABC's base angles at A and B.
The midpoints of AC and BC are D and E respectively.
Segments AD and BE are the "medians drawn from the base angles", and intersect at F.

THE FASTEST PROOF I CAN THINK OF:
The median from vertex C is CG, going through G, the midpoint of AB.

It is also ABC's altitude (perpendicular to) side AB, because ABC is an isosceles triangle.
Point F must be part of median CG, because the medians of a triangle concur.
CF is perpendicular to ABC's base AB, because F is a point in altitude and median CG.

OTHER APPROACHES:
Segments AE, EC, BD, and DC are congruent with one another because D and E are midpoints of congruent sides AC and BC.
Triangles ABE and BAD are mirror-image congruent triangles by SAS (Side-Angle-Side) congruency,
because they have pairs of congruent sides (AE and BD, AB and BA) adjacent to congruent angles at B and A .

Angles BAD and ABE are congruent by CPTC (Corresponding Parts of Congruent Triangles are Congruent).
ABF is an isosceles triangle because it has congruent base angles at A and B.
Triangle ABC's median CG is perpendicular to base AB at midpoint G, because ABF is an isosceles triangle.
Triangle ABF's median FG is perpendicular to base AB at midpoint G, because ABF is an isosceles triangle.
As only one perpendicular to AB can pass through point G, CF and FG must be segments of CG, which is perpendicular to AB at G.

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