請教!中四圓形

Q.1
Since DB//PQ, therefore, angle DBA = angle BAQ = 55 (alternate angles DB//PQ)
Angle BDA = angle BAQ = 55 (angles in alternate segment)
Therefore, angle BDA = angle DBA = 55. Therefore, angle DAB = 180 - 55 - 55 =70 ( angle sum of triangle)
Therefore, x = 180 - 70 = 110 (opposite angle of cyclic quad.)

2008-06-05 17:18:19 補充：
Q.2 Angle ABC = 360 - 40 - 30 - 75 - 75 - 35 = 105.Therefore, angle ABC = angle ADE = 105.Therefore, ABCD is a cyclic quadrilateral (ext. angle equals int. opposite angle).Therefore, angle CBD = angle CAD = 75 (angle in the same segment.)

第一題我上唔到去
我識第2題
求∠CBD
∠D=180度-105度
∠D=75度
180度=∠A+30度+75度
∠A=75度
∠CBD=360度-70度-35度-75度-75度=105度

1.有問題
夾角無理由係同圓周嫁喎~

2.∠CDA=180°-105°=75°­
∠CAD=180°-30°-75°=75°
∠CBD=360°-70°-35°-75°-75°=105°

小朋友，點解你隻∠55度會掂到個circle嫁？