xvnjcke的部落格

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a____________a 中4MATHS 有3條)))))

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a____________a 中4MATHS 有3條))))) [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0006-1.jpg[/IMG]

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最佳解答:

(13a) ∠BDF = ∠BCD (Angle in alt. segment) ∠CBD = ∠DBF (Common angle) So BCD and BDF are similar for the reason of AAA. (13b) ∠BAE = ∠EDC (Alt. angle AB//CD) ∠EFC = ∠EDC (Angle in the same segment) ∠BAE = ∠EFC So, A, B, F and E are concyclic (opp. angles supp) (13c) ∠BAE = ∠EDC (Alt. angle AB//CD) ∠BAE = ∠EDC (Alt. angle AB//CD) (14a) OA and OB are perpendicular to PQ and PR respectively. ∠OAP = ∠OBP = 90° ∠OAP + ∠OBP = 180° So, O, A, P and B are concyclic. (14b) ∠OAQ = ∠OBR = 90° So A, B, R and Q are concyclic for the reason of converse of angle in the same segment. (14c) ∠RAP = ∠QBP = 90° ∠RPA = ∠QPB (Common angle) PA = PB (Tangent properties) So ARP and BQP are congurent for the reason of AAS. (15a) ∠QBP = ∠QCP = 90° ∠QBP + ∠QCP = 180° Therefore, P, B, Q and C are concyclic. (15b) The required circle will have PQ as its diameter since ∠QBP = ∠QCP = 90°. So, radius of the circle = (8 + 5)/2 = 6.5 cm Area = 42.25π cm2

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