1973 AMC 12 Problem 9

Compare the right triangles $CHM$ and $CHB$ using the trisection angles

After locating $M$ and $H$ on $AB,$ compare $HM$ with the whole base $AB$

Since $CH\perp AB,$ both $\triangle CHM$ and $\triangle CHB$ are right at $H.$ The trisection gives equal acute angles at $C,$ and the triangles share side $CH,$ so they are congruent. Hence $HM=HB.$

Because $M$ is the midpoint of $AB,$ $AM=MB.$ Along the hypotenuse the order is $A,M,H,B,$ and $MB=MH+HB=2MH.$ Thus $$AB=2MB=4MH.$$ Triangles $ABC$ and $CHM$ have bases on the same line and the same altitude from $C,$ so their areas are in the ratio $AB:HM=4:1.$ Therefore the area of $\triangle ABC$ is $4K.$

Therefore, the correct answer is E.