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In Fig. 10.39, 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
Solution:
We will use the following concepts to answer the question.

The sum of angles in a triangle is 180°.

Angles in the same segment are equal.
Consider the straightline BD. As the line AC intersects with the line BD, the sum of two adjacent angles so formed is 180°.
Therefore, ∠BEC + ∠DEC = 180°
130° + ∠DEC = 180°
∠DEC =180°  130° = 50°
Consider the ∆DEC, the sum of all angles will be 180º.
∠DEC + ∠EDC + ∠ECD = 180°
50° + ∠EDC + 20° = 180°
∠EDC = 180°  70° = 110°
∴ ∠BDC = ∠EDC = 110°
We know that angles in the same segment of a circle are equal.
∴ ∠BAC = ∠BDC = 110°
☛ Check: NCERT Solutions for Class 9 Maths Chapter 10
Video Solution:
In Fig. 10.39, 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
Maths NCERT Solutions Class 9 Chapter 10 Exercise 10.5 Question 5
Summary:
If 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°, then ∠BAC=110°.
☛ Related Questions:
 ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30° find ∠BCD. Further if AB = BC, find ∠ECD.
 If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
 If the nonparallel sides of a trapezium are equal, prove that it is cyclic.
 Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D, P and Q respectively (see Fig. 10.40). Prove that ∠ACP = ∠QCD.
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