2, 3, 4...
Geometry
Level
4
In triangle ABC, two lines are projected from A and B and meet BC and AC respectively at D and E. These two lines intersect at P, a point in ABC. If the areas of APE, ABP and BDP are 2, 3 and 4 respectively, what is the area of ABC?
The answer is 105.
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1 solution
Draw a line from P to C. Let area of CPE be x and area of CPD = y. Since BPA/APE = 3/2, BP/BE = 3/2 and BPC/CEP = 3/2. Hence (4+y)/x = 3/2 and we have 8 + 2y = 3x ------(1) Since APB/BPD = 3/4, AP/DP = 3/4 and APC/CPD = 3/4 Thus (2+x)/y = 3/4 and we have y = (4/3)(2+x) -------(2) Solving equations (1) and (2), we get x=40 and y= 56. Therefore total area = 2+3+4+40+56 = 105 yay