Question

Suppose AB and CD are different chords of a circle with centre O. Show that if (the extensions of) AB and CD intersect at some point P inside the circle, then ∠APC=1/2(∠BOD+∠AOC)

Answer 1

To show that ∠APC = 1/2(∠BOD + ∠AOC), we can use the fact that angles inscribed in the same arc are equal. By applying this property to the intersecting chords AB and CD in the circle and using congruent triangles, we can derive ∠APC = 1/2(∠BOD + ∠AOC).

Step-by-step explanation:

To show that ∠APC = 1/2(∠BOD + ∠AOC), we can use the fact that angles inscribed in the same arc are equal. Here's how we can prove it:

1. From the given information, we know that AB and CD are chords of the same circle with center O, and they intersect at point P inside the circle.
2. Let ∠AOC and ∠BOD be the central angles of the arcs AC and BD (respectively) that are intercepted by the chords AB and CD.
3. Since angles inscribed in the same arc are equal, we have ∠ABC = ∠ADC and ∠APC = ∠BPD.
4. From triangles ABC and ADC, we have ∠BAC = ∠DAC since they are corresponding angles of congruent triangles.
5. Similarly, from triangles APC and BPB, we have ∠APC = ∠BPD since they are corresponding angles of congruent triangles.
6. Substituting ∠BAC = ∠DAC and ∠APC = ∠BPD into the equation ∠APC = 1/2(∠BOD + ∠AOC), we get ∠DAC = 1/2(∠BOD + ∠AOC).

Therefore, we've shown that ∠APC = 1/2(∠BOD + ∠AOC) using the fact that angles inscribed in the same arc are equal.