Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click on the figure below to enlarge it.
What a wonderful proof!Thank you, Bleaug
thanks for this kind remark...but...my post does not meet the standards of what can be called a 'proof'. I guess you filled in the gaps yourself. So 50% of the proof is yours ;-)bleaug
Αν Α(0,0), Β(Ο,4),C(4,4),D(4,0) και (ε1)=EO: y-2=λ1(x-2) , (ε2)=FO:y-2=λ2(x-2) τότε
<HCO=<BOG=a because <BOC=90. Both triangles HCO and GOB share hypotenuse and the angle a so they are congruent. Similarly OCM and DON are congruent. From those 2 pairs of congruent triangles we have that BG=HO and DN=OM, so the answer is 2(4/sqrt2)^2=16Ivan Bazarov