Consider a circle with centre $O$. Two chords $AB$ and $CD$ extended intersect at a point $P$ outside the circle. If $\angle AOC=43^\circ$ and $\angle BPD=18^\circ$, then the value of $\angle BOD$ is ____?
Assuming that $AB=CD$ I have been able to deduce that it will be $7^\circ$. How to do the sum without the extra assumption?
On
Answer: $\angle BOD=7^o$
Proof: Look at the triangle OAC. Then $\angle OAC=\angle OCA=\frac{137}{2}$.
Now, look at the triangle APC. Then $\angle OAP+ \angle OCP+137+18=180$. This implies that $\angle OAP+ \angle OCP=25^o$.
Look at the quadrilateral ABDC which is inscribed in a circle. Hence $180^o=\angle B+ \angle C=\angle OBA+ \angle OBD+\angle OCP+\frac{137}{2}=$
$\angle OAP+ \angle OBD+\angle OCP+\frac{137}{2}$. This implies that $\angle OBD=180-25-\frac{137}{2}$. Hence $\angle BOD=180-2\angle OBD=7^o$.
Draw $AD$. Then $$ \angle BOD = 2\angle BAD = 2\angle PAD = 2(\angle ADC-\angle APD) = \angle AOC-2\angle APD = 7^\circ $$