Prove that the angle bisectos of a cyclic quadrilateral forms another quadrilateral which is also cyclic.
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21st February 2013,
10:58 PM
Answered by Expert
Answer:
ABCD is a cyclic quadrilateral
?A + ?C = 180 and ?B + ?D = 180
(?A + ?C)/2 = 90 and (?B + ?D)/2 = 90
?A + ?C = 180 and ?B + ?D = 180
(?A + ?C)/2 = 90 and (?B + ?D)/2 = 90
x + z = 90 and y + w = 90
In ?AGD and ?BEC,
x + y + ?AGD = 180 and z + w + ?BEC = 180
?AGD = 180 (x+y) and ?BEC = 180 (z+w)
?AGD + ?BEC = 360 (x+y+z+w) = 360 (90+90) = 360 180 = 180
?AGD+?BEC = 180
?FGH+?HEF = 180
The sum of a pair of opposite angles of a quadrilateral EFGH is 180.
Hence EFGH is cyclic
In ?AGD and ?BEC,
x + y + ?AGD = 180 and z + w + ?BEC = 180
?AGD = 180 (x+y) and ?BEC = 180 (z+w)
?AGD + ?BEC = 360 (x+y+z+w) = 360 (90+90) = 360 180 = 180
?AGD+?BEC = 180
?FGH+?HEF = 180
The sum of a pair of opposite angles of a quadrilateral EFGH is 180.
Hence EFGH is cyclic
Answered by Expert
22nd February 2013,
4:51 PM
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