prove that sum of angle subtended by arcs AC & BD at centre O = twice the angel APC

Question

if two chords AB & CD intersect at P inside the circle

Answer 1

Draw a picture of this situation. Create a circle with center O, and place points A and C near each other on the circle's left half (that is, on its circumference) and points B and D near each other on the circle's right half (on its circumference). The points don't need to be in those positions, it's just easier to picture this if they are.

 

Now, connect A to B (chord AB) and C to D (chord CD). Label their point of intersection P. Finally, draw radii connecting point O to A, B, C, and D.

 

 

<AOC is the angle subtended by arc AC at O, and <BOD is the angle subtended by arc BD at O. Since both are central angles, their measures are equal to their arcs' measures:

 

m<AOC = AC

m<BOD = BD

 

So, the sum of the angles subtended by arcs AC and BD at center O is:

 

m<AOC + m<BOD = AC + BD

 

 

Now, we'll use the formula for the measure of an angle whose vertex lies inside the circle to find m<APC. It involves <APC's vertical angle, <BPD. Note that <APC intercepts arc AC and <BPD intercepts arc BD.

 

angle = (1/2)(angle's intercepted arc + vertical angle's intercepted arc)

m<APC = (1/2)(AC + BD)

 

Since we're interested in twice the measure of <APC, multiply both sides by two:

 

2(m<APC) = AC + BD

 

 

Notice that our two equations now equal the same thing (AC + BD). Set them equal to each other:

 

m<AOC + m<BOD = 2(m<APC)

 

 

Thus, the sum of the angles subtended by arcs AC and BD at center O (<AOC and <BOD) is twice the measure of <APC.