It is proved that if P is passed, PF⊥AC is in F, PH⊥BC is in H, PG⊥AB is in G, and CP is the bisector of △ABC.

∴PF=PH, similarly PH=PG, ∴PF=PH=PG.

Exercise 1 1.3 question 1

Solution: ∵PM⊥OA, PN⊥OB, ∴ OMP = ∠ ONP = 90 in Rt△OMP and Rt△ONP.

OM=ON (known) OP=OP (male * * * side), so Rt△OMP≌Rt△ONP.

So ∠MOP=∠NOP means that OP is the bisector of ∠AOB.

Exercise 1 1.3 Question 6

Solution: The reasons for AD⊥EF are as follows:

Because AD is the bisector of △ABC, ∠BAD=∠CAD, while ∵DE⊥AB, DF⊥AC,

So ∠ AED = ∠ AFD = 90 ∠ In △∠AED=∠AFD △AFD, ∠ bud = ∠ CAD, ∠ AED = ∠ AFD,

Ad = ad，so△aed?△AFD，∴AE=AF.

In △AEG and △AFG, AE = AF, ∠ EAG = ∠ FAG AG = Ag, ∴△AGE≌△AGF.

And ≈age+∠agf = 180, so ∠ age = ∠ AGF = 90, that is, AD ∠ ef.