ABCD is a quadrilateral inscribed in a circle. If angle ADB = 28° and angle CBD = 62°, prove that arc AB + arc CD = arc BC + arc AD. (no diagram is given)
我只係知angle ADB + angle CBD = angle ABD + angle BDC = 90° (opp. angles, cyclic quad.) 但係arc AB + arc CD =/= angle ADB + angle CBD 咁點算??? 要let數???
2)
The diagonals of a cyclic quadrilateral ABCD intersect at K. DA, CB when produced meet at E; AB, DC when produced meet at F. If angle AEB = 55°, angle BFC = 35° and angle DKC = 85°, prove that arc BC is twice arc AB. (no diagram is given)
angle CAB and angle ACB 唔識搵,計到勁多angle都係xx.5°
更新1:
題目冇比圖,係咪要自己估個圖係點?有冇人係唔駛畫返都識做=.=?