证明:(1)连接AF,BG, ∵AC=AD,BC=BE,F、G分别是DC、CE的中点, ∴AF⊥BD,BG⊥AE. 在直角三角形AFB中, ∵H是斜边AB中点, ∴FH=AB. 同理得HG=AB, ∴FH=HG.
(2)∵FH=BH, ∴∠HFB=∠FBH; ∵∠AHF是△BHF的外角, ∴∠AHF=∠HFB+∠FBH=2∠BFH; 同理∠AGH=∠GAH,∠BHG=∠AGH+∠GAH=2∠AGH, ∴∠ADB=∠ACD=∠CAB+∠ABC=∠BFH+∠AGH. 又∵∠DAC=180°-∠ADB-∠ACD, =180°-2∠ADB, =180°-2(∠BFH+∠AGH), =180°-2∠BFH-2∠AGH, =180°-∠AHF-∠BHG, 而根据平角的定义可得:∠FHG=180°-∠AHF-∠BHG, ∴∠FHG=∠DAC. |