The distance between A and BC is equal, and the distance between O and BC and AC is equal, so the distance between O and AB and AC is equal. So 0
About ∠BAC bisector
10. The idea is exactly the same as Question 8. Let p BE the vertical segment of be, BD and AC, and prove that these three vertical segments are in phase.
By analogy, the distance from P to AE side and AC side is equal, and it must be on the angular bisector.
1 1. Extend the CD to point E so that DE=BC.
( 1)∵∠bad+∠BCD = 180 ∴∠abc+∠adc= 180 ∴∠abc=∠ade
In △ABC and △ADE, AB=AD, ∠ABC=∠ADE, BC= DE ∴△ABC≌△ADE.
Ac = AE. ∠ DAE+∠ CAD = ∠ BAC+∠ CAD = 60 ∴△ ACE is an equilateral triangle, ∠ ACE = 60.
Therefore, AC divides ∠BCD equally
(2)△ACE is an equilateral triangle, AC=CE=CD+DE, which has been proved above △ ABC △ ade,
BC=DE, so AC=CD+BC
12. Connect BP and CP. Because p is on the vertical line of BC, BP=CP. P is on the bisector of ∠BAC,
So PE=PF, plus ∠ PEB = ∠ PFC = 90, it can be proved that △ PEB △ PFC. Then BE = CF.