Question

2 In the given AB is the diameter of a circle with centre O. AC and BD are produce meet at E and angle-COD =40 calculate CED.​

Answer 1

Step-by-step explanation:

Given : AB is a diamcter of thc circle with cenlre O, AC and BD produced meet at E and angleCOD= 40". To find : 2CED Solution: join OC & OD now in A AOC OA - OC (Radius) -> ZCAO - ACO-x Simialrly ZDBO = ZBDO =y (as OD = OB = Radius) ABCD is a cylic Quadrilatcral -> ZBAC + ZBDC - 180° (oposile angles) ZBAC = ZOAC as O lies on AB ( as AB is diameler) ZBDC = ZBD0 + ZCOD -> ZOAC ZBDO ZCOD - 180" -> x + y + 2COD - 180 => ZCOD = 180" - x -y ZCOD - ZCDO as (OC - OD - Radius) -> ZCOD - 180° - x - y in AOCD ZDCO - ZCOD + ZDOC= 180" => 2( 180" - x - v) + 40" = 180" => 180" - x - y = 70" in AAEB ZEAB I ZEBA I ZAEB - 180° ZAEB - ZCED => x+y + 2CED = 180' ->ZCED - 180 x y -> ZCED - 70