Question

如圖所示,銳角三角形ABC的內心為I,過點A作直線BI的垂線,垂足為H,點E為圓I與邊CA的切點.

(1)求證A,I,H,E四點共圓;
(2)若∠C=50°,求∠IEH的度數.

Answer 1

(1)見解析   (2) 25°

解:(1)由圓I與AC相切于點E得IE⊥AC,結合HI⊥AH,得∠AEI=∠AHI=90°,所以A,I,H,E四點共圓.
(2)由(1)知A,I,H,E四點共圓,所以∠IEH=∠HAI.由題意知∠HIA=∠ABI+∠BAI=∠ABC+∠BAC=(∠ABC+∠BAC)=(180°-∠C)=90°-∠C,結合IH⊥AH,得∠HAI=90°-∠HIA=90°-(90°-∠C)=∠C,所以∠IEH=∠C.由∠C=50°得∠IEH=25°.