In Fig.10.13, ∠ADC = 130° and chord BC = chord BE. Find ∠CBE.
Solution:
Given, ∠ADC = 130°
Chord BC = chord BE
We have to find ∠CBE
Considering triangle BCO and BEO,
Given, BC = BE
We know that the base angles of equal sides are also equal.
So, ∠BCO = ∠BEO
Common side = BO
By SAS criterion, the triangles BCO and BEO are similar.
By CPCTC,
∠CBO = ∠OBE ------------------------- (1)
From the figure,
ABCD is a cyclic quadrilateral since its points lie on a circle.
We know that the sum of the opposite angles of a cyclic quadrilateral is 180 degrees.
∠ADC + ∠CBA = 180º
130º + ∠CBA = 180º
∠CBA = 180º - 130º
∠CBA = 50º
From the figure,
∠CBA = ∠CBO
So, ∠CBO = 50º
From (1),
∠OBE = 50º
From the figure,
∠CBE = ∠CBO + ∠CBO
= 50º + 50º
= 100º
Therefore, ∠CBE = 100º
✦ Try This: A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and ∠ADC = 130º. Find ∠ABC (in degrees)
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.3 Problem 15
In Fig.10.13, ∠ADC = 130° and chord BC = chord BE. Find ∠CBE.
Summary:
A chord of a circle is any line segment touching the circle at two different points on its boundary. In Fig.10.13, ∠ADC = 130° and chord BC = chord BE, then ∠CBE = 100º
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