Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemClick the figure below to see the complete problem 508 about Triangle with three Squares, Equal Area.
Paralleogram BGTD => ▲BGT = ▲A3 = ▲A1 see P502( ▲M3TG = ▲M3BD, M3 midpoint of DG )In the same way A2 = A1, A4 = A1
A1=area(ABC)=1/2AB.BC.sin(ABC)=1/2.AB.AC.sin(BAC)=1/2.BC.AC.sin(BCA)but sin(ABC)=sin(DBG) (ABC) supplement to (DBG)and sin (BAC)=sin(EAJ)sin(BCA)=sin(FCH)from above we can conclude that A1=A2=A3=A4Peter Tran
Rotate triangle CFH 90 deg. around point C.Point F moves on Point B, H moves to H'.A, C, H' are collinear, and AC=CH', so A1=A4.
Area ABC is (ABC).But <ABC+<DBG=180 so (ABC)/(DBG)=(AB.BC)/(BD.BG)=1 so A1=A3.Similar A1=A2=A4.