The figure is an equilateral triangle. 3 line segments , which meet at a(any) point in the triangle , are of the length 5cm, 4cm, and 3 cm as shown in the figure. Find the side of the equilateral triangle.

  • $\begingroup$ Welcome to Math Stack Exchange. I think we need some more information and for you to reword the question, so that we know what you are asking. Also, there is no figure, did you intend there to be? $\endgroup$ – N. Owad Oct 24 '14 at 16:45
  • $\begingroup$ I am confused as well. Are you saying you have an equilateral triangle with side lengths of $3,4,5$? Because that would not be an equilateral triangle. $\endgroup$ – graydad Oct 24 '14 at 16:51
  • $\begingroup$ @graydad: He's saying that there's a point inside the equilateral triangle, and that the three distances from this point to each of the three vertices are $3$, $4$, and $5$. $\endgroup$ – Lucian Oct 24 '14 at 17:48

Theorem: Let $ABC$ be an equilateral triangle, with a point $P$ inside it such that $PC^2=PA^2+PB^2$, then $\angle APB=150^{\circ}$.

enter image description here

So to do that, we cleverly construct a point $D$ such that $APD=60^{\circ}$ and $AP=PD$. Hence, $APD$ is equilateral and $AP=AD$.

Now, as $60^{\circ}=\angle CAB=\angle DAP$, so, $\angle CAP = \angle BAD$, which is just $\angle PAB$ subtracted from them. We know that $AB=AC$.

So, by SAS criteria $PAC \cong BAD$. Hence, $PC=BD$. This means that $BD^2=PD^2+BP^2$. Thus by converse of Pythagoras Theorem, $\angle BPD = 90^{\circ}$, which means $\angle APB=150^{\circ}$.


Corollary: If $ABC$ be an equilateral triangle, with a point $P$ inside it such that $PC=5$, $PB=4$ and $PA=3$ then $AB=\sqrt{25+12\sqrt3}$.

Note that $3^2+4^2=5^2$, so $\angle APB=150^{\circ}$. Hence, by the cosine law, $AB^2=3^2+4^2-2\cdot12\cdot \cos 150^{\circ}=25+12\sqrt3$. So: $$AB=\sqrt{25+12\sqrt3}$$

  • $\begingroup$ Is that GeoGebra? $\endgroup$ – Soham Chowdhury Oct 24 '14 at 18:05
  • $\begingroup$ @SohamChowdhury Yes :) $\endgroup$ – Sawarnik Oct 24 '14 at 18:07

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