ducand prin G o paralela la FA obtinem punctul D ; deci (GFA)≡(GFAD)
ducand prin D o paralela la BC obtinem punctul A deci(BCD)≡(ABCD)
prin "≡" am intekles identic
(GFAD)∩(ABCD)=AD
AB ⊥AD (ABCD patrat)
AD muchie ⊥ABF, AD ⊥AF⊂(ABF), AF⊥AD
deci m∡((GFAD, (ABCD))=m∡(AF, AB)=45°
