Let $O$ be the centre of the circumcircle of $\Delta ABC$, $P$ and $Q$ be the midpoint of $AO$ and $BC$, respectively.

Question

Let $O$ be the centre of the circumcircle of $\Delta ABC$, $P$ and $Q$ be the midpoint of $AO$ and $BC$, respectively. Suppose $\angle CBA = 4\angle OPQ$ and $\angle ACB = 6\angle OPQ$ . FiNd $\angle OPQ$ .

What I Tried: Here is a picture :-

You can see what I did, I extended some lines and joined them and got a pair of congruent triangles there, and marked some of the angles, and this is my only progress.

Now can anyone help? Thank You.

Answer 1

In your figure, the angle with $4x$ is greater than the angle with $6x$, so $PQ$ is at the wrong side of $O$. Here is a redrawn figure:

The circle is there to aid visualization of the theorem: angle subtended to the circumference is half the angle at center. With this we have:

$$\angle ABC = 4x \implies \angle AOC = 8x, \quad \angle ACB = 6x \implies \angle AOB = 12x$$

Now as $\angle QOB = \angle QOC$, each of them is equal to $\frac12 (360^\circ - 12x - 8x) = 180^\circ - 10x$.

This gives $\angle POQ = 180^\circ - 10x + 8x = 180^\circ - 2x$.

Hence $\angle OPQ = \angle OQP = x$, and $OQ = OP = \frac12OA = \frac12OB$.

By considering $\triangle BOQ$, we see that $\angle BOQ = 60^\circ$. (Why?) Now we can solve for $x$.