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The way three medians of a triangle can form a triangle, whose area is 3/4 the area of the original triangle. The question is can angle bisectors of a triangle form a triangle. I do understand that if we are talking about equilateral triangle, they will because angle bisectors, medians and altitudes are one and the same. What about a scalene triangle, and an iscosecles one.

Medians - I tried using vectors to prove the same. I took the three points and then mid points. Using position vectors, I came up with three vectors whose addition is equal to zero, so I said they will form a triangle. The same thing is not working with angle bisectors.

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  • $\begingroup$ Medians, bisectors, side bisectors and heights, all intersect the point of common intersection is called barycenter, incenter, circumcenter and ortocenter respectively $\endgroup$ – Jorge Fernández Hidalgo Aug 12 '16 at 13:48
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    $\begingroup$ In general, no. Suppose you have an isosceles triangle with two (equal) very long sides and one very short side. Then one bisector is effectively the same as a long side, while the two others are effectively $\sqrt 2$ times the short length. $\endgroup$ – lulu Aug 12 '16 at 15:14
  • $\begingroup$ @CarryonSmiling: I think the question can be simply stated as do the lengths of the angle bisectors fulfill the triangle inequality? Cevians being cevians is quite irrelevant here. $\endgroup$ – Jack D'Aurizio Aug 12 '16 at 17:12
  • $\begingroup$ Your first sentence is not clear at all; it should be such : "in a triangle ABC, with sides' midpoints named $A',B',C'$, vectors $\vec{AA'}\vec{BB'},\vec{CC'}$ have a zero sum (besides, how is it proven ?) and the triangle constitued by putting end to end these vectors has an area 3/4 of the area of the original triangle". $\endgroup$ – Jean Marie Aug 12 '16 at 20:21

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