ΔBOC подобен ΔDOA по 2 углам
\frac{BO}{OD} = \frac{OC}{AO} = \frac{BC}{AD} = \frac{2}{3}
\frac{2AO}{3OC}
AO+OC = 20
2 AO+3OC=40
5OC=40
OC=8
AO=12
ΔBOC подобен ΔDOA по 2 углам
\frac{BO}{OD} = \frac{OC}{AO} = \frac{BC}{AD} = \frac{2}{3}
\frac{2AO}{3OC}
AO+OC = 20
2 AO+3OC=40
5OC=40
OC=8
AO=12