✔ 最佳答案
1)
The centre C of circle ABP lies on circle APQ. Given that ACB and BPQ are straight lines, prove that QC is perpendicular to AB. THE FIRST PHOTO
ㄥABP = ㄥQBC (common)
ㄥCAP = ㄥCQP (ㄥs in the equal segment)
△ABP ~ △QBC (A.A.)
ㄥQCB = ㄥAPB = 90° (∠in semi-circle)
Therefore QC is perpendicular to AB.
2)
The centre O of circle APB lies on circle AQB. OPQ is a straight line. Prove that BP bisects angle ABQ. THE SECOND PHOTO
Let QA meet circle O at A" ,
the mid-point of AA" is a ,
Let QB meet circle O at B"
the mid-point of B"B = b ,
then
ㄥOQB = ㄥOQA (ㄥs in the equal segment)
ㄥOaQ = ㄥObQ = 90° (line joining centre to mid-pt. of chord⊥chord)
QO = QO (common)
△QOa ≡ △QOb (R.H.S.)
So
Oa = Ob
and hence Aa = B"b (chord equidistant from centre are equal)
Therefore
QA = QB"
ㄥOQB = ㄥOQA (ㄥs in the equal segment)
QP = QP (common)
△AQP ≡ △B"QP (S.A.S.)
So AP = B"P
ㄥABP = ㄥQBP (ㄥs in the equal segment)
That means BP bisects angle ABQ.
3)
In a circle ABCD, AB and CD are perpendicular chords intersecting at K. AB=18cm, CD=24cm and the radius of the circle is 13cm. Find the length of KC. (ans corr to 2 d.p.) THE THIRD PHOTO
KC
= DC - DK
= 24 - [ 24/2 + √ (13² - (18/2)²) ]
= 2.62 cm
