In triangle \(ABC\), \(AB=16\), \(AC=24\), \(BC=19\), and \(AD\) is an angle bisector. Find the ratio of the area of triangle \(ABD\) to the area of triangle \(ACD\). (Express your answer as a fraction in lowest terms.)
~ PikachuLovesKetchup
P.S. Thanks!
Well, drawing a diagram will help...
Firstly, once you draw in the angle bisector, you can use the angle bisector theorem to get the ratio of BD to DC.
Secondly, ABD and ACD share the same altitute, so the ratio of their areas is the ratio of their bases. The angle bisector theorem states AB/AC = BD/DC. Since we are currently trying to find BD/DC, we can use AB/AC since we know what AB is and what AC is.
Thus BD/DC = 16/24 = 2/3.
Therefore, the ratio of the area of triangle ABD to the area of triangle ACD is 2/3.
remember the angle bisector theorem, it is helpful...
Thanks! It helped a lot!(I forgot about using theorems literally)
~ PikachuLovesKetchup