In Fig.10.15, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ∠ACD + ∠BED.
Solution:
Given, AOB is a diameter of the circle
C, D and E are any three points on the semicircle
We have to find the value of ∠ACD + ∠BED
Join AE
Since the points A, C, D and E lie on the circle
ACDE is a cyclic quadrilateral
We know that the sum of opposite angles of a cyclic quadrilateral is 180 degrees.
∠ACD + ∠AED = 180° ----------------- (1)
AB is the diameter of the circle
We know that the angle subtended by a diameter to the circle is a right angle
So, ∠AEB = 90° ------------------------- (2)
Adding (1) and (2),
∠ACD + ∠AED + ∠AEB = 180° + 90°
From the figure,
∠BED = ∠AED + ∠AEB
Therefore, ∠ACD + ∠BED = 270°
✦ Try This: ABCD is a cyclic quadrilateral in which ∠BCD = 100° and ∠ABD = 70°.Find ∠ADB.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.3 Problem 19
In Fig.10.15, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ∠ACD + ∠BED.
Summary:
In Fig.10.15, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. The value of ∠ACD + ∠BED is 270°
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