Question

In the figure chord AB and CD when produced meet at point P. If $\angle AOC=\theta ,\angle BOD=\alpha$ then prove that $\angle APC=\frac{\theta -\alpha }{2}$
( O is the centre of the circle )

Answer 1

REF.Image

As exterior angle of a cyclic

quadrilateral is equal to the opposite

interior angle, we have, $\angle PBD=\angle ACD$ and

$\angle PDB=\angle BAC$

Therefore, by AA criteria for symmetry,

$△PBD\sim △PAC$

Let angle $\angle OBA=x=\angle OAB(OA=OB=r)$

and $\angle OBD+\angle ODB+\alpha =\pi$ and $\angle OBD=\angle ODB$

$\Rightarrow \angle OBD=\angle ODB=\frac{\pi }{2}-\frac{\alpha }{2}$; similarly $\angle OAC=\frac{\pi }{2}-\frac{\theta }{2}$

Now, $\angle ABO+\angle OBD+\angle DBP=\pi$

$\Rightarrow x+\frac{\pi }{2}-\frac{\alpha }{2}+\angle DBP=\pi$

$\Rightarrow \angle DBP=\frac{\pi }{2}+\frac{\alpha }{2}-x$

Now, $\angle PBD=\angle BAC$ $(△PBD\sim △PAC)$

$\Rightarrow \frac{\pi }{2}+\frac{\alpha }{2}-x=x+\frac{\pi }{2}-\frac{\theta }{2}\Rightarrow 2x=\frac{\alpha }{2}+\frac{\theta }{2}\Rightarrow x=\frac{\theta +\alpha }{4}$

From symmetry, $\angle PBD=\angle PDB$ and $\angle PBD+\angle PDB+\angle BPD=\pi$

$\Rightarrow 2\left(\angle PBD\right)+\angle BPD=\pi \Rightarrow \angle BPD=\pi -2(\frac{\pi }{2}+\frac{\alpha }{2}-\frac{\theta }{4}-\frac{\alpha }{4})$

$\Rightarrow \angle BPD=\pi -\pi -\alpha +\frac{\theta }{2}+\frac{\alpha }{2}\Rightarrow \angle BPD=\frac{\theta -\alpha }{2}$