In the figure chord AB and CD when produced meet at point P.
If ∠AOC=θ,∠BOD=α
then prove that ∠APC= (θ−α)/2
( O is the centre of the circle )
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To prove that ∠APC = (θ - α)/2, we can use the properties of angles formed by intersecting chords in a circle.
1. Draw a circle with center O and mark points A, B, C, and D as given in the question. Make sure that chords AB and CD intersect at point P.
2. Since O is the center of the circle, O must be equidistant from points A and B. Similarly, O must be equidistant from points C and D. This implies that OA = OB = OC = OD.
3. Since OA = OC, triangle OAC is an isosceles triangle. Therefore, ∠OAC = ∠OCA = θ/2. Similarly, since OB = OD, triangle OBD is an isosceles triangle, and ∠OBD = ∠ODB = α/2.
4. Now, consider the quadrilateral ACBD formed by the intersecting chords. We know that the opposite angles of a cyclic quadrilateral add up to 180 degrees. Therefore, ∠APC + ∠ADC = 180 degrees.
5. ∠ADC is an external angle to triangle ODB, and it can be written as the sum of the two remote interior angles. Thus, ∠ADC = ∠OBD + ∠ODB = α/2 + α/2 = α.
6. Using the sum of angles in triangle ADC, we can write ∠ADC = ∠APC + ∠ACD. Substituting the known values, we have α = ∠APC + θ/2.
7. Rearranging the equation, we find ∠APC = α - θ/2.
8. To simplify the equation, we can multiply both sides by 2 to get 2∠APC = 2α - θ.
9. Finally, we divide both sides by 2 to evaluate the angle ∠APC, resulting in ∠APC = (θ - α)/2.
Therefore, we have proven that ∠APC = (θ - α)/2.