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View diagram here

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

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  • $\begingroup$ Joining $AR$ might help. $\endgroup$
    – RiverX15
    Nov 27 '21 at 7:23
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    $\begingroup$ @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. $\endgroup$ Nov 27 '21 at 8:12
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    $\begingroup$ @Afsheen your diagram is a bit misleading. It depends on whether it is an acute angled isosceles triangle or obtuse angled isosceles triangle. $\endgroup$
    – Math Lover
    Nov 27 '21 at 8:52
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    $\begingroup$ @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. $\endgroup$ Nov 27 '21 at 8:55
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    $\begingroup$ @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. $\endgroup$
    – Math Lover
    Nov 27 '21 at 9:09
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Please see the below diagram which gives a counter-example for obtuse angled isosceles triangle ($120$-$30$-$30$) as mentioned by John Omielan.

enter image description here

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.

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