Mathematics (Quick)－飛航百科

標題:

Mathematics (Quick)

發問:

1. In the figure, ODB and ADC are straight lines. OA=AB=BC. (a) Find angleOBC. (b) Prove that OABC is a parallelogram.http://img391.imageshack.us/my.php?image=14gw9.png2. In the figure, PQ and RS are parallel chords of the circle. PAS and QAR are straight lines. RS is produced to T such that QR=QT. (a) Prove... 顯示更多 1. In the figure, ODB and ADC are straight lines. OA=AB=BC. (a) Find angleOBC. (b) Prove that OABC is a parallelogram. http://img391.imageshack.us/my.php?image=14gw9.png 2. In the figure, PQ and RS are parallel chords of the circle. PAS and QAR are straight lines. RS is produced to T such that QR=QT. (a) Prove that PS//QT. (b) Prove that ST=PQ. http://img176.imageshack.us/my.php?image=15kg0.png 3. In the figure, AB and CD are two parallel chords. Chords BC and AD intersect at E. (a) Prove that AE=BE. (b) Prove that triangleABE ~ triangleCDE. http://img176.imageshack.us/my.php?image=16ky2.png 4. In the figure, BC is a diameter of the circle and ABC is a right-angled triangle with angleABC = 90degree. E and D are points on AB and AC respectively such that ED is the tangent to the circle at D. (a) Prove that EB=ED. (b) Prove that E is the mid-point of AB. http://img147.imageshack.us/my.php?image=17uk2.png

最佳解答:

1a) As OA = OB = AB (OA = OB since they are radii of circle) △OAB is an equil. triangle and therefore ∠OAB = ∠OBA = ∠AOB = 60 Hence ∠ACB = 30 (Angle at centre = Twice angle at circumference) With AB = BC, △BAC is an isos. triangle with ∠BCA = ∠ACB = 30 So ∠ABC = 180 - 30 - 30 = 120 and finally, ∠OBC = ∠ABC - ∠OBA = 60 (b) Since ∠OBC = ∠BOA = 60, AO // BC (Converse of alt. ∠s) With AO = BC given, we have OABC being a parallelogram. 2a) ∠PQR = ∠QPR = ∠QRT = ∠PSR for the reason of angles in the same segment and alt. ∠s PQ // RS. So ∠QTR = ∠QRT as △QRT is an isos. triangle and hence ∠QTR = ∠PSR. Finally, PS // QT. (b) With △APQ ~ △ARS (AAA) and AQ = AP and AR = AS (as they are isos. triangles) we have: AQ + AR = AP + AS QR = PS Therefore PS = QT and thus PQTS is a parallelogram. Finally, PQ = ST. 3a, b) ∠EBA = ∠EAB = ∠ECD = ∠EDC for the reason of angles in the same segment and alt. ∠s AB // CD. So △EBA is isosceles with AE = BE and also △ABE ~ △CDE (AAA) 4a) Join OD and then ∠ODE = 90 since radius is perp. to tangent. So ∠ODE = ∠OBE. OB = OD (Radii of circle) OE = OE (Common side) So △OBE and △ODE are congurent (RHS) and therefore EB = ED. (b) With OB = OD and ∠ODE = ∠OBE = 90, we have OBED is a square. Also BC = 2OB since BC is a diameter of the circle. So BC = 2DE and hence E is the mid-point of AB according to the mid-point theorem.