binwangbin80
幼苗
共回答了17个问题采纳率:88.2% 举报
连PD.
∵AB=AC,AD⊥BC,∴BD=CD,∠BAD=∠CAD.
显然有:∠PDC>∠ADC=∠ADB>∠PDB.
由BD=CD,PD=PD,∠PDC>∠PDB,得:PC>PB,∴∠PBC>∠PCB.
由AB=AC,得:∠ABC=∠ACB,结合证得的∠PBC>∠PCB,
得:∠ABC-∠PBC<∠ACB-∠PCB,即:∠PBA<∠PCA.
而∠PAB<∠BAD=∠CAD<∠PAC,∴∠PBA+∠PAB<∠PCA+∠PAC,
∴180°-(∠PBA+∠PAB)>180°-(∠PCA+∠PAC),
又∠APB=180°-(∠PBA+∠PAB),∠APC=180°-(∠PCA+∠PAC),
∴∠APB>∠APC.
1年前
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