2. As shown in the figure, in △ABC,? D is a moving point on the side of AC, extending AB to E, making BE=CD, connecting de, and intersecting BC at point P. Verify that DP=PE.
3.。 As shown in the figure, in △ABC, ∠B=2∠C, and AD divides ∠BAC equally. Verification: AC=AB+BD.
4. It is known that two separate triangles △ABC and △XYZ are acute triangles, AB=XY and BC=YZ.
∠C=∠Z, verification: △ABC? ≌? △XYZ (no graph)
1、
Prove:
∫△ABC and△△△ DCE are equilateral triangles.
∴BC=AC,CD=CE,∠ACB=∠DCE=60
∴∠ACB+∠ACD=∠DCE+∠ACD
Namely: ∠ BCD = ∠ ACE.
∴△BCD≌△ACE(SAS)
∴∠BDC=∠AEC
Also ∵B, C, E three-point * * * line
∴∠fcd= 180-∠ACB-∠DCE = 60 =∠DCE
∴△FCD≌△GCE(ASA)
∴CF=CG
2、
Prove:
Point D is DF‖AB, and point BC is point F.
∫△ABC is a regular triangle.
∴∠CDF=∠A=60，∠CFD=∠CBA=60，∠C=60
∴∠CDF=∠CFD=∠C=60
△ CDF is a regular triangle.
∴CD=DF
Cd = become again
∴DF=BE
Ni DF‖AB
∴∠PDF=∠PEB
DPF = BPE again
∴△DFP≌△EBP(AAS)
∴DP=PE
3、
Prove:
Intercept AE = AB on AC and connect DE.
Advertising split ∠BAC
∴∠BAD=∠DAE
Ad = ad again.
∴△ABD≌△AED(SAS)
∴∠B=∠AED,BD=DE,AB=AE
∫∠B = 2∠C
∴∠AED=2∠C
∠ AED =∠ C +∠ EDC。
∴∠EDC=∠C
∴DE=CE
∴BD=CE
∴AC=AE+EC=AB+BD
4、
Prove:
Pass through point B as BD⊥AC in D.
YD'⊥XZ is at d' if you cross point Y.
∠ BDC =∠ yd 'z = 90。
∠ c =∠ z，BC = yz。
∴△BCD≌△YZD'(AAS)
∴BD=YD'
And ab = xy, ∠ADB =∠xd'y = 90.
∴Rt△ABD≌Rt△XYD'(HL)
∴∠A=∠X
∠ c =∠ z，ab = xy。
∴△ABC？ ≌? △XYZ(AAS)