[tex]AOD=\frac{1}{3} \times DOC=\frac{1}{4} \times COB[/tex]
Stim ca:
∡AOD+∡DOC=∡AOC
∡AOC=180°-∡BOC
Inlocuim si vom obtine:
[tex]\frac{1}{3}\times DOC+DOC=AOC\\\\ \frac{4}{3}\times DOC=AOC[/tex]
Din ipoteza avem:
[tex]\frac{1}{3} \times DOC=\frac{1}{4} \times COB\\\\BOC=\frac{4}{3}\times DOC[/tex]
∡AOC=180°-∡BOC
Vom inlocui cele de mai sus si vom obtine:
[tex]\frac{4}{3}\times DOC=180- \frac{4}{3}\times DOC[/tex]
Aducem la acelasi numitor comun:
4×∡DOC=540-4×∡DOC
8×∡DOC=540°
∡DOC=67,5°
∡COB=4×67,5:3=90°
∡AOD=67,5:3=22,5°