9514 1404 393
Answer:
ON : NB = 5 : 4
Explanation:
It may be helpful to refer to the attached figure.
We have divided the triangle into pieces so that we can refer to specific areas and sums of areas. For example, the area of triangle AMO is a+b1. That is half the total area, due to segment AM being half the total length AB.
We have found it convenient to choose a total area of 126 square units. The labels in the figure show how that total area is divided among the different triangles.
Here's one way to get there.
a + b1 = 1/2 the total, so is 63
Similarly,
d + c + b2 = 63
The given segment ratio OP : PM = 5 : 2 tells us ...
a : b1 = 5 : 2 ⇒ a = 45, b1 = 18
Since PM is the median of ∆ABP, we have ...
b1 = b2 = 18 ⇒ d + c = 45
So far, we know
d + c = ∆BOP = 45 and ∆BOA = 126
which means AN : PN = 126 : 45 = 14 : 5.
Applying that ratio to triangles to the right of AN, we have ...
c / (c + b1 + b2) = 5 : 14
14c = 5(c +36) ⇒ c = 20
Since c + d = 45, this means d = 25.
The ratio of d to c is the ratio of ON to NB:
d : c = 25 : 20 = 5 : 4 = ON : NB
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Additional comment
We have made extensive use of the fact that triangles with the same base have areas proportional to their height. In the final conclusion, we have also used the fact that triangles with the same height have bases proportional to their areas.