图1性质如图1,点P是△ABC的内心,过点P垂直于AP的直线分别交AB、AC于点D、E,则DE是△PBC外接圆的切线.证明∵点P是△ABC的内心,DE⊥AP,显然易证Rt△APD≌Rt△APE,∴∠ADE=∠AED,在△ADE中,∠ADE+∠AED+∠DAE=180°,即2∠ADE=180°-∠DAE①同理∠ABC+∠ACB=180°-∠BAC②由①、②得∠ADE=12∠ABC+12∠ACB,而∠ADE=∠DBP+∠DPB=12∠ABC+∠DPB,∴∠DPB=12∠ACB=∠PCB, Figure 1, the nature of Figure 1, the point P is the center of △ ABC, the point P perpendicular to the AP of the line were AC, AC at the point D, E, then DE is △ PBC circumcircle tangent. In the center of ABC, DE ⊥ AP, it is clear that it is easy to prove Rt Δ APD ≌ Rt Δ APE, ∴∠ADE = ∠ AED. In Δ ADE, ∠ADE + ∠AED + ∠DAE = 180 °, ie 2∠ADE = 180 ° -∠ DAE ① empathy ∠ ABC + ∠ ACB = 180 ° -∠BAC ② from ①, ② was ∠ADE = 12∠ABC + 12∠ACB, and ∠ADE = ∠DBP + ∠DPB = 12∠ABC + ∠DPB, ∴∠DPB = 12∠ ACB = ∠PCB,