Question

In the figure given below, O is the centre of the circle and AB is a diameter.
If AC = BD and ZAOC = 72º. Find:
1. Angle ABC
2. Angle BAD
3. Angle ABD

Answer 1

Answer:

AO=CO(radii)

in isosceles triangle AOC,

AOC+OAC+ACO= 180

2OAC=180-72

OAC=BAC=54

therefore, BAC=54

Given, AC=BD

ACB=ADB=90(angle formed by diameter and a point at the circumference=90°)

hence in traingle ABC,

ABC+BAC+ACB=180

ABC=180-(54+90)

ABC=36

in triangle ABC and BAD

ACB=BDA=90

AC=BD

AB=AB(common)

Therefore ∆ABC is congruent to ∆BAD(RHS)

ABC=BAD(cpct)

BAC=ABD(cpct)

BAD=36

ABD=54