1. In the figure, AOB is a diameter and BC is tangent to the circle at B. Line AEC meets the circle at D such that E is the mid-pt of AD. If OB=BC,
(a) find angle BEC,[Ans.45 degree]
(b) prove that DE=DB.
http://i226.photobucket.com/albums/dd260/wingreally/0031e93d.jpg
2. The figure shows the circumcircle of triangle ABC. BM meets AC at E at right angles while AL meets BC at D at right angles. BM and AL meet at N. Prove that
(a) angle LBD= angle NBD,
(b) ND=DL,
(c) DE parallels with LM, LM =2DE.
http://i226.photobucket.com/albums/dd260/wingreally/1-1.jpg
3. In the figure, circles O and O' touch each other at P and PE is their common tangent. APB and ACED are straight lines. AD is tangent to the circle O' at D. Prove that
(a) angle CPD= angle DAB+ angle ABD,
(b) PD bisects angle BPC,
(c) PD^2= (PC)(PB).
http://i226.photobucket.com/albums/dd260/wingreally/2-1.jpg
4. In the figure, AC, AB and BC are the diameters of semi-circles ADC, AEB and BFC respectively. AD cuts the semi-circle AEB at E while DC cuts the semi-circle BFC at F . DB is perpendicular to AC. Prove that
(a) EF and BD are equal and bisect each other,
(b) EF is a common tangent to semi-circles AEB and BFC,
(c) AEFC is a cyclic quad.
http://i226.photobucket.com/albums/dd260/wingreally/3.jpg
Thx~
Maths about circle
2009-08-09 8:41 am
回答 (1)
2009-08-09 4:04 pm
✔ 最佳答案
Q1.For triangle ADB, angle ADB = 90 degree ( angle in semi-circle).For triangle ABC, angle ABC = 90 degree ( CB is tangent of circle).
Triangle ADB is similar to triangle ABC (AAA, angle A is common)....(1)
OB = BC (given)
OB = OA (radius of circle)
therefore, AB = OA + OB = 2BC.
Based on result of (1), AD = 2DB = AE + ED = 2ED
so DB = DE.
Therefore, triangle EDB is a right - angled isos. triangle, so
angle DEB = (180 - 90)/2 = 45 degree.
2009-08-09 08:22:11 補充：
Q2. Triangle BEC similar to triangle ADC (AAA, angle C is common), so angle CBE = angle DAC.......(1). Angle LBC = angle DAC ( angle in same segment), so angle CBE = angle LBC. Therefore, triangle DBL congruent triangle NBD ( ASA, BD is common), so ND = DL.
2009-08-09 08:25:05 補充：
Q2 cont'd. Angle EAM = angle EBC ( angle in same segment), so angle NAE = angle EAM, so triangle MAE congruent triangle EAN ( ASA), so NE = EM and ND = DL (proved), so DE//LM and LM = 2DE ( mid-point theorem).
2009-08-09 08:35:52 補充：
Q3. Angle CAP = angle CPE and angle DPE = angle PBD ( both reason is angle in alt. segment), so angle CAP + angle PBD = angle CPE + angle DPE = angle CPD.
2009-08-09 08:38:43 補充：
Q3 cont'd. Angle CDP = angle PBD ( angle in alt. seg.). angle PAC + angle CDP = angle DPB ( ext. angle of triangle), so angle DPB = angle CPD, so PD is the angle bisector of angle CPB.
2009-08-09 08:40:29 補充：
Q3 cont'd. Based on the above results, triangle CPD is similar to triangle DPB (AAA). So PC/PD = PD/PB, that is PD^2 = PC x PB.
2009-08-09 08:53:09 補充：
Q4. Angle AEB = angle ADC = angle BFC = 90 degree ( angle in semi-circle) So BEDF is a rectangle, so EF = DB and bisects each other ( property of rectangle).
2009-08-09 08:56:44 補充：
Q4 cont'd. Let EF intersects BD at M. Since DB is tangent to circle AEB and angle EBM = angle MEB ( property of rectangle), so EM and MB are tangents to circle AEB. Similarly, BM and MF are tangents to circle BFC, so EMF = EF is the common tangent.
2009-08-09 09:00:58 補充：
Q4 cont'd. Angle BEF = angle BDC ( property of rectangle). Angle BDC + angle DCB = 90 degree ( angle sum of triangle DBC). So anfle BEF + angle DCB = 90 degree, so angle BEF + angle DCB + angle AEB = 180 degree. So AEFC is a cyclic quad. ( oppo. angle complement).
收錄日期: 2021-04-25 22:36:59
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