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Question 5
In the given figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that BEC=130 and ECD=20. Find BAC.



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Solution


Given,
BEC=130
ECD=20
BEC+DEC=180 [Linear pair]
130+DEC=180
DEC=180130=50

Now, in ΔDEC
DEC+ECD+CDE=180 [Angle sum property of a triangle]
50+20+CDE=180
CDE=18070=110
CDE=BAC [Angles in the same segment]
BAC=110


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