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 $PQ \parallel BC$?

  • $\begingroup$ Joining $AR$ might help. $\endgroup$
    – RiverX15
    Nov 27 '21 at 7:23
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.