Question

Two chords $AB$ and $CD$ intersect at $P$ inside the circle. Prove that the sum of the angles subtended by the arcs $AC$ and $BD$ at the centre $O$ is equal to twice the angle $APC$.

Answer 1

Reffer image.

we know angle subtended by an arc of a circle at the centre is twice the angle subtended at any part of circle .

$\angle AOD=2\angle ABD.....\left(i\right)$

$\angle BOC=2\angle BDC.....\left(ii\right)$

adding (i)and(ii)

$\angle AOD+\angle BOC=2\angle ABD+2\angle BDC.....\left(iii\right)$

$\angle APD=\angle BDP+\angle DBP.....\left(iv\right)$ (exterior of a triangle = sum of opp. interior angles)

from the figure,

$\angle PBD=\angle ABD$ and $\angle PDB=\angle CDB$

$\therefore$ from (iii) and (iv)

$\angle AOD+\angle BOC=2\angle APD$