Question 388372: Given: ABCD is arhombus.
prove:AC bisects angle BAD and angel BCD.
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website! To prove this statement we need to use the result that the diagonals of a parallelogram (which the rhombus is a particular kind of) bisect each other. Let the point of intersection of AC and BD be E. Then since the sides of a rhombus are congruent, triangle AEB is congruent to traingle AED, and so by congruence of corresponding parts, angle BAC is congruent to angle DAC. By similar reasoning, it can be shown that angle BCA is congruent to angle DCA.
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