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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
Solution:
We will use the following concepts to answer the question:

A quadrilateral ABCD is called cyclic if all the four vertices of it lie on a circle.

The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.

The sum of angles in a triangle is 180°.

Angles in the same segment are equal.
Based on the data given, let's draw the figure as shown below.
In the triangles, ABD and BCD, ∠CAD = ∠CBD = 70°. (Angles in the same segment are equal)
Hence, ∠BAD = ∠CAB + ∠DAC
= 30° + 70° = 100°
Thus, ∠BAD = 100°
Since ABCD is a cyclic quadrilateral, the sum of either pair of opposite angles of a cyclic quadrilateral is 180º.
∠BAD + ∠BCD = 180°
∠BCD = 180°  100°
= 80°
Thus, ∠BCD = 80°
Also given AB = BC.
So, ∠BCA = ∠BAC = 30° (Base angles of isosceles triangle are equal)
∠ECD = ∠BCD  ∠BCA
= 80°  30°
= 50°
Thus, ∠ECD = 50°
☛ Check: Class 9 Maths NCERT Solutions Chapter 10
Video Solution:
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
Maths NCERT Solutions Class 9 Chapter 10 Exercise 10.5 Question 6
Summary:
If ABCD is a cyclic quadrilateral whose diagonals intersect at a point E, ∠DBC = 70°, ∠BAC = 30°, and AB = BC, then ∠BCD = 80° and ∠ECD = 50°.
☛ Related Questions:
 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.
 If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
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