Question

四边形ABCD中,AD,BD的延长线交于E,AB,DC的延长线

Answer 1

⑴for ∠a+∠afd+∠adc=180°

∠adc=136°-∠afd

for ∠bcd=∠afd+∠cbf=136°

∠cbf=136°-∠afd

so ：∠cbf=∠adc

⑵ for ∠cbf=∠adc,∠pfc=∠pfa,∠peb=∠ ped,∠bcf=∠dce

2∠pfc+∠cbf=136°

∠adc=∠dce+2∠peb so 2(∠peb+∠pfc) +∠dce=136°

for ∠dce=180°-∠bcd=44°

so ∠peb+∠pfc=46°

⑶ ∠epf=180°-(∠a+∠pfa+∠ped)=180°-（64°+∠peb+∠pfc）=70°