F4 圓形特性 Geometry - Circle

For triangle ADH and triangle AHC
AD = AC (given)
angle DAH = angle HAC ( AQ is the angle bisector of angle A)
AH = AH (common)
therefore they are congruent (SAS)
so angle DHA = angle AHC = right angle = angle CHQ
For triangle AQC
since angle CHQ = right angle (proved)
angle ARC = right angle (given)
therefore, K is the orthocenter of triangle AQC
so QK is perpendicular to AC.
Let QK meets AC at X
so triangle AQX and triangle AHC are similar (AAA)
therefore, angle AQK = angle ACH.
so angle QKH = angle CAH ( triangle AHC similar triangle KHQ)...(1)
Since BP = PC (given) and
DH = HC ( proved)
therefore PH // AB ( mid- point theorem)
therefore angle PHQ = angle BAH ( corres. angle PH//AB)........(2)
since angle BAH = angle CAH ( AQ is the bisector)........(3)
From (1), (2) and (3) we get
angle QKH = angle PHQ
therefore PH is a tangent to circle HKRQ ( alternate angle equal)
Q.E.D.






2009-06-17 11:20:26 補充：
To be exact, reason for the last line is " angle in alt. segment equal'.