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ABCD是一圓內接四邊形，E、F、G和H分別是△DAB、△ABC、△BCD和△CDA的內心。證明EFGH是一矩形。

連AE、AF、BE、BF。

∠EAF=∠EAB-∠FAB=(1/2)(∠BAD-∠BAC) (內心特性)
∠EBF=∠FBA-∠EBA=(1/2)(∠CBA-∠DBA) (內心特性)

ABCD是一圓內接四邊形
∴∠ADB=∠ACB (同弓形內的圓周角)
∴∠ABC+∠CAB=180°-∠BCA (三角形內角和)
∴∠ABC+∠CAB=180°-∠ADB
∵180°-∠ADB=∠BAD+∠ABD (三角形內角和)
∴∠ABC+∠CAB=∠BAD+∠ABD
∴∠BAD-∠BAC=∠CBA-∠DBA
∴∠EAF=∠EBF
∴A、B、F和E共圓 (同弓形內的圓周角的逆定理)

∴∠EFB=180°-∠BAE (對角互補)
∴∠EFB=180°-(1/2)∠BAD

同理，∠GFB=180°-(1/2)∠DCB
∠BAD+∠DCB=180° (對角互補)
∠EFB+∠GFB=180°-(1/2)∠BAD+180°-(1/2)∠DCB=360°-(1/2)(∠BAD+∠DCB)=360°-(1/2)(180°)=270°
∴∠EFG=360°-(∠EFB+∠GFB)=360°-270°=90° (同頂角)

同理可證，∠EFG=∠FGH=∠GHE=∠HEF=90°
∴EFGH是一矩形

1000 for siuhochung

Thanks

same site

http://www.kalva.demon.co.uk/

IMO has Q&A website
How about APMO?  

you know where it is. APMO 96 A3

Can post the exact source?  

Take the points as A, B, C, D in that order. Let I be the incenter of ABC. The ray CI bisects the angle ACB, so it passes through M, the midpoint of the arc AB. Now angle MBI = angle MBA + angle IBA = angle MCA + angle IBA = (angle ACB + angle ABC)/2 = 90 deg - (angle CAB) /2 = 90 deg - (angle CMB)/2 = 90 deg - (angle IMB)/2. So the bisector of angle IMB is perpendicular to IB. Hence MB = MI. Let J be the incenter of ABD. Then similarly MA = MJ. But MA = MB, so the four points A, B, I, J are concyclic (they lie on the circle center M). Hence angle BIJ = 180 deg - angle BAJ = 180 deg - (angle BAD)/2.

Similarly, if K is the incenter of ADC, then angle BJK = 180 deg - (angle BDC)/2. Hence angle IJK = 360 deg - angle BIJ - angle BJK = (180 deg - angle BIJ) + (180 deg - angle BJK) = (angle BAD + angle BDC)/2 = 90 deg. Similarly, the other angles of the incenter quadrilateral are 90 deg, so it is a rectangle.

這是以前APMO的題目∼

圓內接四邊形ABCD，證明ABC,BCD,ACD,ABD的內心組成一個長方形∼
(這題應該不會有座標-平面幾何之爭吧∼)