Question 246805
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The problem as stated cannot be done.  You cannot prove something that in general is not true.  Now if MATH were actually a rhombus, that is another story altogether.


Here's the proof that your Prove statement is false in general.


Construct two parallelograms that are on top of one another:  *[tex \LARGE MAT_1H_1] and *[tex \LARGE MAT_2H_2] such that *[tex \LARGE MH_2] is significantly greater than *[tex \LARGE MH_1].  Construct the diagonals *[tex \LARGE MT_1] and *[tex \LARGE MT_2].


Since angle *[tex \LARGE AMH_1] is congruent to *[tex \LARGE AMH_2] but the alleged half-angles *[tex \LARGE AMT_1] and *[tex \LARGE AMT_2] are clearly of different measures, the two diagonals cannot be bisectors.  



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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