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Let $\Delta ABC$ be an isosceles triangle, s.t. $\overline{AC}$ and $\overline{BC}$ have the same length. Let $D$ be some point on $\overline{BC}$, s.t. $\overline{BD}$ and $\overline{CD}$ have the same length. Assume that $\Delta ABD$ is also an isosceles triangle, s.t. $\overline{AB}$ and $\overline{AD}$ have the same length.

Find the size of the angle $\angle ACB$. (Marked red in the sketch)

Sketch of the triangle

My idea to make use of the assumption $\left| \overline{BD} \right| = \left| \overline{CD} \right|$ was to add a copy of the triangle $\Delta ABD$ to the outside of $\overline{CD}$. One receives a parallelogram $ABA'C$, but I cannot seem to make use of it. If I recall correctly, an elementary solution exists and trigonometry is not necessary, but I have failed to find it for much time now...

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  • $\begingroup$ see Apollonius theorem (en.wikipedia.org/wiki/Apollonius%27_theorem) $\endgroup$ – Jean Marie Apr 7 '17 at 19:58
  • $\begingroup$ That shows $\frac{1}{2} \left| \overline{AC} \right|^2 = \left| \overline{AD} \right|^2$. From this I can compute the angle using the law of cosines, that's right. Thank you! I was hoping to find a solution without trigonometry though... $\endgroup$ – user425553 Apr 7 '17 at 20:15
  • $\begingroup$ I wasn't informed of your answer because you hadn't put @ my pseudo $\endgroup$ – Jean Marie Apr 7 '17 at 20:17
  • $\begingroup$ @user425553 Is it possible that $AD=CD$ and not $BD = CD$? Because using an alternative method I also got scarface's answer and $\arccos\left(\frac{3}{4}\right)$ is approximately $41.41^{\circ}$ which is not an easy angle to get. $\endgroup$ – Futurologist Apr 8 '17 at 4:34
  • $\begingroup$ @Futurologist: I also realized that $arccos \left( \frac{3}{4} \right)$ cannot be deduced from the elementary theorems on angles. Your guess may be correct and the exercise is supposed to state $AD=CD$ instead of $BD=CD$, as the picture suggests. In which case the angle would be $36^\circ$. $\endgroup$ – user425553 Apr 8 '17 at 8:28
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Answer: Exact form is $\angle ACB = \arccos \dfrac34$ and approximate form is $\angle ACB \approx 41,4^\circ$.

Let's solve it. Without loss of generality we can assume that $|CD|=|BD|=1$, $|AC|=2$ and $|AD|=|AB|=x$. Since $\triangle ACB \sim \triangle BAD$, $\dfrac2x= \dfrac x2$ and so $x=\sqrt 2$. By cosine theorem in $\triangle ABC$, $$\cos \angle ACB =\dfrac{2^2+2^2-2}{2\cdot 2\cdot 2}=\dfrac 34$$.

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