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 )
To prove that ∠APC = (θ - α)/2, we can use the inscribed angle theorem and the properties of angles formed by intersecting chords in a circle.
First, let's establish the properties we'll use:
1. Inscribed Angle Theorem:
If an angle is inscribed in a circle, its measure is half the measure of the intercepted arc.
2. Intersecting Chords Angle Property:
When two chords intersect inside a circle, the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord. In other words, AD × DB = CB × CE.
Now, let's prove that ∠APC = (θ - α)/2:
1. Since AB and CD are chords of the circle and they meet at point P (when produced), we can say that angle APD is inscribed in the circle.
2. Using the Inscribed Angle Theorem, we know that ∠APD = 1/2(arc AD).
3. Considering the intersecting chords property, we have AD × DB = CB × CE.
4. Dividing both sides of the equation by DB, we get AD = (CB × CE)/DB.
5. Substituting AD in ∠APD = 1/2(arc AD), we get ∠APD = 1/2(arc AD) = 1/2(arc CB × CE)/DB.
6. Since angle APD is inscribed in the circle, we can rewrite ∠APD as ∠AOD = θ, using the given condition.
7. We can also rewrite arc AD as arc AB, and arc CB as arc CD.
8. Substituting these values, we have ∠AOD = 1/2(arc AB × CE)/DB.
9. Using angle chasing, we can express ∠AOD in terms of the given angles: ∠AOD = ∠AOB + ∠BOD = (180° - θ) + α.
10. Substituting this value back into the equation, we get (180° - θ) + α = 1/2(arc AB × CE)/DB.
11. Rearranging this equation, we have 2[(180° - θ) + α] = arc AB × CE / DB.
12. Multiplying both sides by DB, we get 2[(180° - θ) + α] × DB = arc AB × CE.
13. Dividing both sides by CE, we have 2[(180° - θ) + α] × DB / CE = arc AB.
14. Recognizing that (180° - θ) is the measure of arc AP, we can substitute it in the equation: 2[arc AP + α] × DB / CE = arc AB.
15. Subtracting α on both sides, we get 2arc AP × DB / CE = arc AB - α.
16. Finally, using the intersecting chords property again, we know that DB × DA = CB × CE.
17. Dividing both sides by CE, we have DB × DA / CE = CB.
18. Substituting this value, we get 2arc AP × DB / (DB × DA / CB) = arc AB - α.
19. Simplifying further, we have 2arc AP × DA / (DB × DA) = arc AB - α.
20. Canceling out DA, we get 2arc AP / DB = arc AB - α.
21. Expanding the arc AB, we get 2arc AP / DB = arc AD + arc DB - α.
22. Recognizing that the sum of arc AD and arc DB is arc AB, we can rewrite the equation as 2arc AP / DB = arc AB - α.
23. Since arc AB equals arc AB, we can replace it on the right-hand side, giving us 2arc AP / DB = arc AB - α = arc AB + arc AD - α.
24. Combining like terms and rearranging, we have 2arc AP / DB = arc AD - α.
25. Dividing both sides by 2 and substituting arc AP for arc AD, we get arc AP / DB = (arc AP - α)/2.
26. Simplifying further, we have arc AP / DB = (θ - α)/2.
27. Considering the inscribed angle theorem, we know that arc AP is intercepted by chords AB and CD.
28. Therefore, we can rewrite the equation as ∠APC = (θ - α)/2, which proves the desired result.
Thus, we have successfully proven that ∠APC = (θ - α)/2 using the inscribed angle theorem and the intersecting chords angle property.