In the figure given below, 0 is the centre of the circle and AB is a diameter. °
IfAC = BD and LAOC = 72 . Find: (i) LABC
(ii) LBAD
(iii) LABD
Answers
Step-by-step explanation:
Recall some properties of circle:
1. Angle at the centre of a circle is twice the angle at the circumference
2. In an isosceles trapezoid, the upper base angles are congruent. The lower base angles are also congruent.
3. Triangle inscribed in a semicircle is a right angle triangle.
4. Sum of the angles in a triangle is 180.
Find∠ABC:
\text {Reason: Angle at the centre of the circle is twice the angle at the circumference}Reason: Angle at the centre of the circle is twice the angle at the circumference
\angle ABC = \angle AOC \div 2∠ABC=∠AOC÷2
\angle ABC = 72 \div 2∠ABC=72÷2
\angle ABC = 36∠ABC=36
Find∠BAD:
\text{Reason: The upper base angles of an isosceles trapezoid are congruent }Reason: The upper base angles of an isosceles trapezoid are congruent
\angle BAD = \angle ABC∠BAD=∠ABC
\angle BAD = 36∠BAD=36
Find∠ABD:
\text{Reason: triangle inscribed in a semicircle is a right angle triangle}Reason: triangle inscribed in a semicircle is a right angle triangle
\angle ADB = 90∠ADB=90
\text {Reason: Sum of angles in a triangle is 180}Reason: Sum of angles in a triangle is 180
\angle ABD = 180 - 90 - 36∠ABD=180−90−36
\angle ABD = 54∠ABD=54