Question

如图所示,锐角三角形ABC的内心为I,过点A作直线BI的垂线,垂足为H,点E为圆I与边CA的切点.

(1)求证A,I,H,E四点共圆;

(2)若∠C=50°,求∠IEH的度数.

Answer 1

解:(1)由圆I与AC相切于点E得IE⊥AC,结合HI⊥AH,得∠AEI=∠AHI=90°,所以A,I,H,E四点共圆.

(2)由(1)知A,I,H,E四点共圆,所以∠IEH=∠HAI.由题意知∠HIA=∠ABI+∠BAI=∠ABC+∠BAC=

(∠ABC+∠BAC)= (180°-∠C)=90°-∠C,结合IH⊥AH,得∠HAI=90°-∠HIA=90°-

(90°-∠C)=∠C,所以∠IEH=∠C.由∠C=50°得∠IEH=25°.