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In the following diagram of a triangle, $\overline{AB} = \overline{BC} = \overline{CD}$ and $\overline{AD} = \overline{BD}$. Find the measure of angle $D$.

I know this should be easy but I am stuck. I started by saying angle $\widehat{ACB} = \theta$ and that the supplement angle $\widehat{BCD} = 180^\circ-\theta$. I know that angle $\widehat{CAB}=\theta$ as well and that angle $\widehat{ABC} = 180^\circ-2\theta$. In addition, angles $\widehat{CDB}$ and $\widehat{CBD}$ are equal. I am not sure how to solve for angle $\widehat{CDB}$ ... is it possible to find an exact numerical measure? I hate overlooking something obvious. Thank you for your help.

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  • $\begingroup$ Did you use the fact that the triangle are isoceles? $\endgroup$ – wesssg Sep 17 '17 at 23:42
  • $\begingroup$ There are three isosceles triangles: $\Delta ABC$ and$\Delta ADB$ are similar, but also $\Delta BCD$ is isosceles.The solution drops out of this. $\endgroup$ – Lubin Sep 17 '17 at 23:46
  • $\begingroup$ I suggested an edit to your post using MathJax. If you don't like the specific notation, you can change it of course, but you should still use MathJax $\endgroup$ – gen-z ready to perish Sep 17 '17 at 23:54
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\begin{align} 5\,\delta&=180^\circ \end{align}

Edir

\begin{align} |AB|&=|BC|=|CD| ,\\ |AD|&=|BD| . \end{align}

Let $\angle BDA=\delta$.

Then, from isosceles $\triangle BDC$, $\angle CBD=\delta$, $\angle DCB=180^\circ-2\,\delta$.

In $\triangle ABC$, $\angle BCA=180^\circ-\angle DCB=2\,\delta$, $\angle BAC=\angle BCA=2\,\delta$.

Also, $\angle BAC=\angle BAD=\angle ABD$, hence $\angle ABD=2\,\delta$.

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  • $\begingroup$ I'm afraid I don't fully understand your response. You have 5 d (delta?) = 180. I understand ABD is an Isosceles triangle but the method by which you develop 2 delta for all the angles in triangle ACB, I am thoroughly stumped. Please clarify these old neurons. Thank you. I am trying to understand the process so I can use it for an exam coming up. $\endgroup$ – James Price Sep 18 '17 at 0:34
  • $\begingroup$ @James Price: clarification added. Good luck with the exam. $\endgroup$ – g.kov Sep 18 '17 at 0:50
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$\triangle ABD$ is isosceles with each base angle double the angle at the vertex. The angle at D is therefore $36^o$. Euclid constructs this triangle in Elements,IV,10, as necessary for construction of the regular pentagon.

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  • $\begingroup$ Ok, I looked up on Google the basis of your answer found here: mathcs.clarku.edu/~DJoyce/java/elements/bookIV/propIV10.html I even read through their explanation. Is there any way of knowing exactly for sure that this is 36 -72-72 as opposed to say some other measure. Is it that the angle must always be double the vertical angle? Does this mean as well that all angles in this are either 72 or 36 as answer? Thank you for your explanation and help! $\endgroup$ – James Price Sep 18 '17 at 0:41
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    $\begingroup$ Since it's given that triangles $ABD$ and $ABC$ are isosceles, and situated as indicated, we can see they are similar. And since we know the exterior angle ACB equals the sum of the interior angles CDB +CBD, then angle ACB must equal twice angle CDB alone. And angle ACB equals the base angles DAB and DBA of the big isosceles triangle. This seems to uniquely determine triangle ABD as having angles, 36-72-72. Does that seem sufficient? $\endgroup$ – Edward Porcella Sep 18 '17 at 5:27

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