# $PQ ∥ BC$ for isosceles $\triangle ABC$ and inscribed equilateral $\triangle PQR$ with $R$ being midpoint of $BC$

Triangle $$ABC$$ is isosceles. An equilateral triangle $$PQR$$ is inscribed in it with $$R$$ being the midpoint of $$BC$$. How can you prove $$PQ \parallel BC$$?

• Joining $AR$ might help. Nov 27 '21 at 7:23
• @MathLover Actually, it's not necessarily true $BP = CQ$ based on just the information provided. Using the law of sines in $\triangle BRP$ and $\triangle CQR$ shows that $\sin(\measuredangle CQR) = \sin(\measuredangle BPR)$, so either $\measuredangle CQR = \measuredangle BPR$ or $\measuredangle CQR + \measuredangle BPR = 180^{\circ}$. In the former case, you are correct but, in the latter case, one can show that $\measuredangle PBR = \measuredangle QCR = 30^{\circ}$, with it then being possible to have $PB \neq CQ$ and $PQ \not\parallel BC$. The diagram then would appear quite different. Nov 27 '21 at 8:12
• @Afsheen your diagram is a bit misleading. It depends on whether it is an acute angled isosceles triangle or obtuse angled isosceles triangle. Nov 27 '21 at 8:52
• @MathLover Yes, with an acute angled isosceles triangle, your comment is correct. Otherwise, as my previous comment indicates, if $\measuredangle BAC = 120^{\circ}$ (i.e., is an obtuse angled isosceles triangle), then a counter-example is possible. Nov 27 '21 at 8:55
• @JohnOmielan yes you are right. I did not notice that the triangle was not acute angled. I just added a diagram that shows the counter-example. Nov 27 '21 at 9:09

Please see the below diagram which gives a counter-example for obtuse angled isosceles triangle ($$120$$-$$30$$-$$30$$) as mentioned by John Omielan.

For $$\angle BRP = 60^\circ + x, \angle CRQ = 60^\circ - x$$ or vice versa with $$0 \lt x \lt 30^\circ$$ will give us points $$P$$ and $$Q$$ on sides $$AB$$ and $$AC$$ such that $$\triangle PQR$$ is equilateral but $$PQ$$ is not parallel to $$BC$$.

By law of sines, we can show that $$120$$-$$30$$-$$30$$ is the only isosceles triangle for which $$PQ$$ is not necessarily parallel to $$BC$$.

Say $$\angle B = \angle C = y$$ and $$\angle BRP = 60^\circ + x, \angle CRQ = 60^\circ - x$$

By law of sines in $$\triangle BPR$$,

$$\displaystyle \frac{\sin (180^\circ - (60^\circ + x+y))}{BR} = \frac{\sin y}{PR} \tag1$$

By law of sines in $$\triangle CQR$$,

$$\displaystyle \frac{\sin (180^\circ - (60^\circ - x + y))}{CR} = \frac{\sin y}{QR} \tag2$$

As $$BR = CR$$ and $$PR = QR$$, from $$(1)$$ and $$(2)$$ we obtain

$$\sin (60^\circ - x + y) = \sin (60^\circ + x + y)$$

So we either have $$60^\circ - x + y = 60^\circ + x + y ~$$ i.e. $$~x = 0$$. That leads to $$\angle BRP = \angle CRQ = 60^\circ ~$$ and $$~PQ \parallel BC$$.

Or we have,

$$(60^\circ - x + y) + (60^\circ + x + y) = 180^\circ \implies y = 30^\circ$$ and $$\triangle ABC$$ is $$120$$-$$30$$-$$30$$ triangle. In this case, it is not necessary that $$\angle BRP = \angle CRQ$$. I have demonstrated this case in the first part of my answer.

• That's an excellent diagram and explanation of the issue I mentioned in my comments. Nov 27 '21 at 9:09
• Nov 27 '21 at 15:47