If the angle MPN is known, ad is on PM, C and B are on PN, and a and B intersect CD on F. (1) if pf bisects the angle MPN, the proof is: 1 / PA + 1 / Pb = 1 / PC + 1 / PD; (2) if 1 / PA + 1 / Pb = 1 / PC + 1 / PD, PF bisects the angle MPN

If the angle MPN is known, ad is on PM, C and B are on PN, and a and B intersect CD on F. (1) if pf bisects the angle MPN, the proof is: 1 / PA + 1 / Pb = 1 / PC + 1 / PD; (2) if 1 / PA + 1 / Pb = 1 / PC + 1 / PD, PF bisects the angle MPN


① It is proved that extending BP to point e makes PA = PE, connecting AE. Extending DP to point G makes PC = PG, connecting CG. Because pf bisection angle MPN, ∠ DPF = 1 / 2 ∠ DPB = g, so PF ‖ CG is the same as PF ‖ AE, and DG = PD + PG = PD + pcbe = Pb + PE = Pb + PA, so PF / GC = Pd / DG = Pd / (PD + PC) ① P



It is known that in △ ABC, ad bisects ∠ BAC, EF bisects vertically, and the intersection of AD and BC extension line is at point E


∵ point E is on the vertical bisector of segment ad, ∵ EA = ed,
∴∠EAD=∠EDA,
∵∠EAD=∠EAC+∠CAD,∠EDA=∠B+∠DAB
And ∵ ∠ CAD = ∠ DAB,
∴∠EAC=∠B.



As shown in the figure, ad is the bisector of ∠ EAC, ad ‖ BC, ∠ B = 30 ° to calculate the degree of ∠ DAC, ∠ C


∫ ad ∥ BC, ∠ B = 30 °. ∫ ead = ∠ B = 30 ° (two straight lines are parallel and the same angle is equal). ∫ ad is the bisector of ∠ EAC, ∫ DAC = ∠ ead = 30 °. ∫ ad ∥ BC, ∫ C = ∠ DAC = 30 ° (two straight lines are parallel and the same angle is equal)



As shown in the figure, it is known that ad is the bisector of ∠ EAC, and ad ‖ BC


It is proved that ∵ ad ∥ BC, ∵ ad ∥ BC, ∵ 1 = B, ∵ 2 = C. ∵ ad bisection ∵ EAC, ∵ 1 = 2. ∵ B = C