Synchronous training 1

1.⑴C； ⑵C。

2.⑴AC = a 1c 1； ⑵CE，△ABF?△CDE。

3. prove △ Abe △ ace.

4. Connect BC to prove △ ABC △ DCB.

5.( 1) Prove △ ade △ CBF; (2) prove ∠AEF=∠CFE.

6.( 1) AE=CF or AF=CE can be added to prove △ dec △ BFA; 5] BFA = ∠ DEC, ∴DE∥BF starts from (1).

Synchronous training 2

1.⑴A； ⑵A； ⑶B。

2.⑴∠COB、SAS、CB； ⑵BAD，CAD，AB=AC，∠BAD=∠CAD，AD=AD，SAS。

3. certificate △ ABC △ ade.

4. Divide and prove △ ABC△ ADC.

5. The answer is not unique. There are two options: (1) choose 2 from 134; ⑵ From ① ② ④ to ③, the proof is abbreviated.

6.( 1) AC ⊥ CE, certificate △ ABC △ CDE; This conclusion still stands.

Synchronous training 3

1.⑴C； ⑵A； ⑶B。

2.( 1) AB = CD or OA=0C or ob = od Υ AAS, AB, DC, AAS, △ Abe, △ DCE; ⑶①②③⑤

3.∵∠ 1=∠2, ∴∠BAC=∠DAE, and ∵AC=AE, ∠C=∠E, ∴△ABC?.

4. Prove △ AEF △ CED.

5. Get ∠ADB=∠AEC from ∠ 1=∠2, and then prove △ Abd △ ace with AAS.

6.⑴△AOE?△AOD，△BOE?△COD，△AOB?△AOC，△ABD?△ACE； (2) correct; (3) For example, we can first prove that △ AOE △ AOD gets OE=OD, and then prove that △ BOE △ COD gets BE=CD.

7.( 1)∫ab = a 1b 1，∠ADB =∠a 1d 1b 1 = 90，∴△ADB?△a 65438。 (2) If △ABC and △ A1B1are both acute triangles or both right triangles or both obtuse triangles, then AB=A 1B 1 BC=B 1Cl, ∠. Then △

Synchronous training 4

1.⑴C； ⑵D； ⑶C。

2.⑴ 16; ⑵DCF，HL。

3. Because AE⊥BC and DF⊥BC are in Rt△ABE and Rt△DCF, Rt△ABE≌Rt△DCF, so ∠ ABC = ∠ DCB.

4. Because CE=BF, CE+EF=BF+EF, that is, BE=CF, in Rt△AEB and Rt△DCF, so △ Abe △ DCF, so ∠B=∠C, so AB∨CD.

5. Because DF⊥AC, DE⊥AB, ∠ Bed = ∠ CFD = 90. In △BDE and △CDF, so △ BDE △ CDF, so DE = DF.

In Rt△AED and Rt△AFD, so Rt△AED≌Rt△AFD, so ∠BAD=∠CAD, that is, AD shares ∠BAC.

6.B。