Question

in the adjoining figure, o is the centre of the circle and AB is the diameter. if AB parallel CD and angle ABC =25 DEGREE find angle CED.

Answer 1

CD is the chord of the circle, and the angle subtented between the lines drawn from the ends of the chord to a point on the circle (E in this case) is always constant no matter where E lies, but E should be on the circle.

Angle subtented by a diameter of the circle onto a point on the circle is 90 degree.

Hence Angle ACB is 90, Angle ABC is 25.

So since the sum of the angles in a triangle is 180 degrees we can find that

Angle CAB is 65 degree.

Now move E to A, we can reposition it as I said before. So Triangle ABC is the same as Triangle EBC since right now A and E are the same points.

Now if we see carefully Triangle ABC and Triangle ABD are mirror images of each other since they are similar triangles.

So we can say that Angle BAD is equal to angle ABC which is 25 degrees.

Also Now we can see that

Angle CAB = Angle CAD + Angle BAD

Angle CAD = 65-25
Angle CAD = 40 degrees.
Angle CAD = Angle CED

So angle CED is 40 degrees.

Answer 2

∠CED = 40° Chord CD is parallel to diameter AB & Angle ABC = 25

Step-by-step explanation:

CD ║ AB

=> ∠BCD = ∠ABC = 25° (Alternate angles)

Join CO and DO.

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by an arc at any point on the circumference.

=> ∠BOD = 2∠BCD

=> ∠BOD = 2 × 25° = 50°

Similarly, ∠AOC = 2∠ABC

=>  ∠AOC = 2 × 25° = 50°

AB is a Diameter hence passing through the centre.

=> ∠AOC + ∠COD + ∠BOD = 180°

=>  50° + ∠COD + 50° = 180°

=>∠COD  = 80°

∠CED=∠COD /2

=> ∠CED= 80°/2 = 40°

=>∠CED = 40°

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