Prove that all medians of a triangle are concurrent.

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excuse me should we prove it in coordinates or vectors

2015-04-21T23:58:03+05:30
median joins a vertex to the midpoint of the opposite side. The diagram shows all three medians which are concurrent at a point called the centroid.
if u want to prove it with a figure then draw as attatched figure.

required to prove :-
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//ABProof: 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//ABClaim: Angle DGE = angle AGBProof : 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.

hence all medians are congruent

plzzz mark the stars beside thanks below the answer if it help u

2015-04-22T00:27:27+05:30

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See diagram.

In a triangle ABC, the points D, E and F are midpoints of the sides.

Compare the two triangles ABC and AFE.  The sides AF || AB  and  AE || AC. The angle A is common.  The ratio of sides  AF/AB = AE/AC = 1/2.  Then the two triangles AEF and ABC are similar (due to side-angle-side ratio property).

The angles AFE and ABC are equal.  Then AEF and ACB are also equal.  Hence, EF and AB are parallel and  EF = AB/2.

Let the medians BE and CF intersect  at O.  We need to prove that AO and OD are parallel.

Look at the triangles  OGE and ODB.
OB || OE  and  GE || BD .         The included angles  GOD and DOB (vertically opposite angles) are same.  The angles  GEO and EBD are equal, alternate angles between two parallel lines.   Similarly,  the angles OGE and ODB are equal (alternate angles).

The two triangles are similar.

Hence,  OD  || OG.    ie., OD  || OA.  Hence proved.