Midpoint


From Gogeometry Using Pythagoras central and inscribed angle Equilateral triangle DC=DA and ∠ADC= π/3 => ΔADC is equilateral∠CAD= π/3, ∠BAC+∠CAD=∠BAD=75° => ∠BAC=15°∠ACD= π/3, ∠BCA+∠ACD=∠BCD=155° => ∠BCA=75°∠BAC=15° et ∠BCA=75° => ∠ABC= π/2 ΔADC is equilateral and EC=ED => ∠AEC= π/2∠ABC= π/2 et ∠AEC= π/2 => A, B, C and E are […]

Gogeometry Problem 405


From Gogeometry Using Additional points Pythagoras congruent triangles Similar triangles Equilateral triangle Let b=AD=DC and d=BCΔACB and ΔBCD are similar since ∠BAC=∠DBC=x and ∠BCD=∠BCAThen AC/BC=CB/CD=AB/BD => 2b/d=d/b => d=b√2 Define C’ such as DC’ ⊥ AC and DC’=b∠BDC=∠ADC – π/4= π – π/4 = 3 π/4∠BDC’=∠BDA + ∠ADC’ = π/4 […]

Gogeometry Problem 001