AB=3AD
Sabcd=AB*AD
Sabcd=3AD*AD=3AD²
AD=√Sabcd/3
AD=√12/3=√4=2cm
AB=2×3=6cm
Pabcd=2AB+2AD=2×6+2×2=16cm
AT║OB
DO=OB(O la intersectia diag)
BC=CT
Atunci OB=linie mijlocie in Δ ATC
Atunci AT║OB
ΔTGC=isoscel(BC=BT si GC=inaltime si mediatoare dec si bisectoare a unghiului TGC)
Atunci∡BGC=∡BGT
In Δ BGT,
∡BGT+∡BTG=90°
Dar inΔTOC.
∡BTG+TCO=90°
Atunci ∡BGT=∡TCO
InΔABC, dreptunghic in B,
tga=AB/BC=6/2=3; Am notat cu a=∡TCO
Deci a=arctg3