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 andBCD, ∠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 ABCDis 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 non-parallel 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.
