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Fri 5 Aug, 2005 04:53 pm
Given triangle ABC. Let E be a point on line AB and D be a point on line AC such that line DE is parallel to line BC and BD bisects angle ABC. If AE=4, AC=18 and BC=15 then what is the length of DC?
Since BC is parallel to DE, angle CBD is equal to angle BDE (alternate interior angles). Therefore triangle BDE is isosceles, and BE=DE.
Triangles ABC and AED are similar, so BC/DE=AB/AE.
BC=15, AE=4, AB=4+BE, DE=BE
Substituting, we get 15/BE=(4+BE)/4
You can now solve for BE, and the rest is just proportions.
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