In triangle ABC, the bisector AD intersects the median BM at point O, with BO: OM = 4: 3.

In triangle ABC, the bisector AD intersects the median BM at point O, with BO: OM = 4: 3. What is the ratio of the areas of triangles DBO and ABC?

Consider a triangle ABC.

So AD is a bisector, then AO is a bisector. Consider a triangle ABM.

By the property of the bisector of a triangle for ABM we have:

AB / AM = BO / OM = 4/3.

Then AM = 3/4 * AB and

AC = 2 * AM = 3/2 * AB. Means: AB / AC = 2/3.

By the property of the bisector of a triangle for ABC we have:

AB / AC = BD / DC = 2/3.

Now consider the BMC triangle.

Area of ​​a triangle DBO:

S1 = 0.5 * BO * BD * sin (OBD).

Triangle area MCB:

S2 = 0.5 * BM * BC * sin (OBD).

Area of ​​triangle ABC:

S = 2 * 0.5 * BM * BC * sin (OBD) = BM * BC * sin (OBD), because BM is the median.

S1 / S = 0.5 * BO * BD * sin (OBD) / BM * BC * sin (OBD) = 0.5 * BO * BD / BM * BC.

But BM / BO = (BO + OM) / BO = 1 + OM / BO = 1 + 3/4 = 7/4 and

BC / BD = (BD + DC) / BD = 1 + DC / BD = 1 + 3/2 = 5/2.

BO / BM = 4/7, BD / BC = 2/5.

S1 / S = 0.5 * BO * BD / BM * BC = 0.5 * 4/7 * 2/5 = 4/35.