因为 ∠AA1B1=∠AA1C1=90∘\angle AA_1B_1=\angle AA_1C_1=90^\circ∠AA1B1=∠AA1C1=90∘,点 B1,C1B_1,C_1B1,C1 在以 AA1AA_1AA1 为直径的圆上。在该圆中,AC1=AA1sin∠AA1C1=AA1sin(90∘−∠A1AC)=AA1sinCAC_1=AA_1\sin\angle AA_1C_1=AA_1\sin(90^\circ-\angle A_1AC)=AA_1\sin CAC1=AA1sin∠AA1C1=AA1sin(90∘−∠A1AC)=AA1sinC,同理 AB1=AA1sinBAB_1=AA_1\sin BAB1=AA1sinB,B1C1=AA1sinAB_1C_1=AA_1\sin AB1C1=AA1sinA;因此三角形 AC1B1AC_1B_1AC1B1 与三角形 ABCABCABC 相似。