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Six answers to autonomous learning mathematics
8. Make vertical line segments of AB, AC and BC from point O, respectively. According to BD and CE are bisectors, we can get O to.

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.