Point K is chosen on the median of an isosceles triangle ABC such that BK:KM = 1:2. Find the ratio of the areas of triangles ABK and ABC.

Question

Point K is chosen on the median of an isosceles triangle ABC such that BK:KM = 1:2. Find the ratio of the areas of triangles ABK and ABC.

 

Answer 1

My guess is 1 to 6 or Area of ABK * 6 = Area of ABC.

 

Six triangles of equal area are created when the centroid X is drawn.

 

Triangle BAM is exactly one half the area of ABC.

 

Triangle AXM is 1/6 of the total area where X is the centroid .

 

So AXB is 1/3 of the total area.

 

 

By construction, MX =KX = BK  because the medians

force a 2 to 1 ratio and are of equal length for an

isosceles triangle. So AMX and AKX have equal area.

 

 

So 1/2 of the total area - 2 x 1/6 of the total area = 1/2 - 1/3 of the total area = 1/6 of the total area

= area of AKB.

 

Diagram available, but it's sloppy.