AD bisects ∠BAC, ②

DE⊥AB,DF⊥AC,③

AD ⊥ ef. Taking two of these three as conditions and the other as conclusion, three propositions can be formed, namely: ① ②.

③,①③

②,②③

①.( 1) Try to judge whether the above three propositions are correct (answer directly); Please prove the proposition that you think is correct. See if it is the answer to this question: (1)①②.

3. Correct; ①③

2. Error; ②③

(1), correct. (2) Evidence comes first

①②

③ As shown in figure 1.

Divide ∠BAC, DE⊥AB, DF⊥AC equally in AD, and AD

=

AD,∴

Rt△ADE≌Rt△ADF,∴

Delaware

=DF，∠ADE

= ∠ ADF。 Let AD and EF intersect at g, then △ deg △ dfg, so ∠DGE.

=∠DGF, then ∠DGE.

=∠DGF

=

90 years, so advertising ef. Re-certification.

②③

① As shown in Figure 2, let the midpoint of AD be O and connect OE, of.

∵

DE⊥AB,DF⊥AC,∴

OE and OF are the median lines of the hypotenuse of Rt△ADE and Rt△ADF, respectively.

That is, the distance from point O to A, E, D and F is equal, so the four points A, E, D and F are centered on O,

On a circle with rADius, ad is the diameter. So EF is a chord of ⊙O, and EF⊥AD, ∴.

AD split equally, that is, so ∠DAE.

=∠DAF, that is, AD shares ∠ BAC.