B bisects AC (why?), and so, the bases of triangles and are each half the length of , which is also a similar triangle. Since these are similar, corresponding sides are proportional, and knowing is double the magnitude of the two other triangles, we know the exact proportion. The upshot of all this is:
since AP and CQ are corresponding sides to AR with half its magnitude.
Now I spouted a lot of unproven claims here. It's up to you to fill in the blanks
construct two perpendicular lines to AC, one passing through P, the other through Q. That will give you the sine relationships your professor mentioned (Remember SOHCAHTOA!) The fact that the sum is 1/2 AC goes back to what I was saying about similar triangles with the factor of 1/2
A point on a line can be placed anywhere if there is no spec given.To draw a diagram for a proof lets make AB= 4 cm BC = 6 cm AC =10cm. Points P, Q, R lie on the perpendicular bisectors of AB,BC, AC respectively. Three 30-30-120 isosceles triangles are erected as described. Extend AP and QC to meet the perpendicular bisector of AC at S above AC
ARCS and PBQS are parallelograms AP+QC = AR.The whole is the sum of the parts
sorry if i am being a moron ,but ....
suppose AP and QC when extended meet as S. Can we be sure that the perpendicular bisector of AC always passes through the point S ???Originally Posted by bjhopper
another doubt: is the original trig solution ok?