SOLUTION: Given: Triangle ABC with AD bisecting angle BAC, and AE is congruent to ED (both line segments). The question is asking for me to prove that AE/AC = BD/BC.
The picture has an
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-> SOLUTION: Given: Triangle ABC with AD bisecting angle BAC, and AE is congruent to ED (both line segments). The question is asking for me to prove that AE/AC = BD/BC.
The picture has an
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Question 474056: Given: Triangle ABC with AD bisecting angle BAC, and AE is congruent to ED (both line segments). The question is asking for me to prove that AE/AC = BD/BC.
The picture has an isoceles triangle with "A" at the top and the base being "B" on the left, and "C" on the right. The bisector is AD and "D" is between "B" and "C" on the base of this triangle. There is also a segment drawn from "D" to "E". a point on the AC segment on the right side of the triangle. (Sorry - not sure how to get the symbols)!
Thanks for your help! Answer by richard1234(7193) (Show Source):
First, we note that triangle ADE is isosceles, so angle ADE equals alpha. Now we can say that lines DE and BA are parallel, so triangles CDE and CBA are similar (by an AAA argument).
Because they're similar, we can establish
Rewrite CD as CB - BD and CE as CA - AE.
, as desired. Here I switched the order of the points (e.g. BC to CB) because each segment refers to the magnitude of the line segment. There are some theorems where line BC might refer to a vector quantity, in which BC = -CB, and switching the order of the points would change everything.
