Prove that the meridian of triangle all congruent
Answers
Answer:
Step-by-step explanation:
The medians of a triangle are concurrent and the point of concurrence, the centroid, is one-third of the distance from the opposite side to the vertex along the median.
Proof: Given triangle ABC and medians AE, BD and CF. So F is the midpoint of AB, E is the mipoint of BC and D is the midpoint of AC by definition of the median.
First, draw medians AE and BD and segment DE.
Claim: Triangle ABC is similar to triangle DEC.
Proof: Angle ACB = angle DCE; AC = 2CD; BC = 2CE; so similar by Side-Angle-Side Similarity Theorem.
Claim: DE//AB
Proof: Angle CDE = Angle CAB and Angle CED = Angle CBA from similarity of triangles ACE and DCE.
Claim: Angle GED = angle GAB and angle GDE = angle GBA.
Proof: DE//AB
Claim: Angle DGE = angle AGB
Proof : Vertical interior angles are congruent.
Thus triangle ABG is similar to triangle EDG by the Angle-Angle-Angle Similarity Theorem.
Thus DE/AB = GE/GA =1/2, which implies GE = 1/2 GA, which in turn implies GE = 1/3 AE. Also GD = 1/2 GB, which implies GD = 1/3 BD by the above.
Repeating the above for AE paired with CE and BD paired with CE, we see that each pair intersects at a point that cuts each median into two pieces at a point such that the piece closest to a side has 1/2 the length of the piece closest to the vertex. That point can be only one point and that is G.
