P23, question 6:

AD⊥EF。

Prove:

In δ δ△AED and δ△AFD:

∠EAD=∠FAD (definition of angular bisector)

∠AED=∠AFD=90

AD=AD (male side)

∴△AED≌△AFD(AAS)

∴∠ADE=∠ADF。

∫ad is the angular bisector of df⊥ac de⊥ab△ABC,

∴DE=DF (the distance from the point on the bisector of the angle is equal to both sides of the angle),

∴DG is the bisector of the vertex angle of isosceles δδcef,

∴AD vertical EF.

P27 .

Question 5: In RT δ BDE and RT δ CDF,

bd=cd,be=cf,∴δbde≌δcdf,∴de=df，

∵DE⊥AB,DF⊥AC，

∴AD bisector ∠BAC, (points with equal distance to both sides of the angle are on the bisector of this angle)

That is, AD is the angular bisector of δδABC.

Question 6: Draw bisectors (straight lines) of the inner and outer angles of a triangle respectively. * * * There are four intersections, and all four points meet the conditions.

The attached drawings are being sent. ?