性质1如图1,任意△ABC中,AD⊥BC,D为垂足,P是AD上距垂足近的三等分点,以P为圆心,PA为半径的圆交BC或其延长线于点M、N.则△AMN是等边三角形.证明连结PM,∵P是AD的三等分点,1∴PD=PA,2而PA=PM,1∴PD=PM,2在Rt△PDM中,显然∠PMD=30°,从而∠MPD=60°,∵∠PMA=∠PAM,1∴∠PMA=∠MPD=30°,2∴∠AMN=∠PMA+∠PMD=60°, Property 1 As shown in Fig. 1, for any △ ABC, AD ⊥ BC, D is the foot of the foot, P is the third equinox near AD from the foot, P is the center of the circle, PA is the radius of the circle BC or its extension At point M, N, then ΔAMN is an equilateral triangle. Prove that the link PM, ∵P is the third-degree point of AD, ∴PD = PA, 2 and PA = PM, 1∴PD = PM, 2 in Rt △ PDM, it is clear that ∠ PMD = 30 °, so ∠MPD = 60 °, ∵∠PMA = ∠PAM, 1∴∠PMA = ∠MPD = 30 °, 2∴∠AMN = ∠PMA + ∠PMD = 60 °,