prove that all medians of a triangle are concurrent.
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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 DCF; AC = 2CD; BC = 2CF; so similar by Side-Angle-Side Similarity Theorem.Claim: DE//ABProof: Angle CDE = Angle CAB and Angle CFD = Angle CBA from similarity of triangles ACE and DCE.Claim: Angle GED = angle GAB and angle GDE = angle GBA.Proof: DE//ABClaim: Angle DGF = angle AGBProof : Vertical interior angles are congruent.Thus triangle ABG is similar to triangle FDG by the Angle-Angle-Angle Similarity Theorem.Thus DF/AB = GF/GA =1/2, which implies GF = 1/2 GA, which in turn implies GF = 1/3 AF. Also GD = 1/2 GB, which implies GD = 1/3 BD by the above.Repeating the above for AF 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.
Hence Proved
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 DCF; AC = 2CD; BC = 2CF; so similar by Side-Angle-Side Similarity Theorem.Claim: DE//ABProof: Angle CDE = Angle CAB and Angle CFD = Angle CBA from similarity of triangles ACE and DCE.Claim: Angle GED = angle GAB and angle GDE = angle GBA.Proof: DE//ABClaim: Angle DGF = angle AGBProof : Vertical interior angles are congruent.Thus triangle ABG is similar to triangle FDG by the Angle-Angle-Angle Similarity Theorem.Thus DF/AB = GF/GA =1/2, which implies GF = 1/2 GA, which in turn implies GF = 1/3 AF. Also GD = 1/2 GB, which implies GD = 1/3 BD by the above.Repeating the above for AF 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.
Hence Proved
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