4

The problem is as follows:

In the figure. Find $x$.

Sketch of the problem

The choices given in my book are:

$\begin{array}{cc} 1.16^{\circ}\\ 2.18^{\circ}\\ 3.20^{\circ}\\ 4.\frac{45}{2}^{\circ}\\ 5.15^{\circ}\\ \end{array}$

According to the official answers sheet the answer for this problem is choice 5. But how to get there?

So far the only clue which I found is that the lines $AD$ and $BC$ are parallel but what else can it be done here?. How to solve this without requiring trigonometry?.

Is there some sort of construction needed? Can someone help me here please?.

Chris Steinbeck Bell
  • 4,175
  • @WillJagy, $AB$ and $DC$ aren't parallel as $\langle BAC \neq \langle ACD$ – PC1 Jun 21 '22 at 01:36
  • @PC1 you are right. I named one of the angles $t$ for angle ABD, so that $t + 6x = \pi.$ Then I filled in all the angles, there is no further requirement. Take any angle $x$ with $6x < \pi...$ I see, ix $x$ increases to $30^\circ$ the diaIgram blows up. If $x = 15^\circ$ we get a right angle at ABD. Meanwhile, by alternate angles, AD really is parallel to BC – Will Jagy Jun 21 '22 at 02:44
  • In brief, angle $x$ can be anything above $0^\circ$ and below $30^\circ$ – Will Jagy Jun 21 '22 at 02:45

1 Answers1

3

enter image description here First of all, since alternate angles $\angle CAD=\angle ACB=x$, we have $AD\parallel BC$. Therefore, $\angle CBD=\angle ADB=3x$.

Next, we construct $\triangle BCE\cong\triangle BCD$ as shown in the figure above. We then have $\angle CBD=\angle CBE=3x$. Notice that $A$, $B$, and $E$ are on the same line, since $\angle CBE=\angle BAC+\angle BCA$.

Let's assume $AB=BD=BE=a$, and $AD=CD=CE=b$.

In $\triangle BCE$, $\angle CBE=3x$ and $\angle BCE=2x$. Meanwhile, in $\triangle ACE$, $\angle ACE=3x$ and $\angle CAE=2x$. This implies $$\triangle ACE\sim\triangle CBE$$ and therefore $$\frac{AE}{CE}=\frac{CE}{BE}\iff \frac{2a}{b}=\frac{b}{a}\iff b=\sqrt{2}a$$

We conclude that $\triangle ABD$ is an isosceles right triangle, so $\color{red}{x=15^{\circ}}$.

rogeryen
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  • Your answer is interesting but I wonder, is it possible to solve this problem without using triangle similarity?. I say this because this problem belongs to a chapter in my book where such concept is not yet introduced and I believe it can be solved using maybe congruency or some other construction. So, can you suggest some alternate option please?. – Chris Steinbeck Bell Jun 23 '22 at 08:15
  • @ChrisSteinbeckBell what book? – Will Jagy Jun 23 '22 at 23:32
  • @WillJagy My book just says Geometry Maths 2, it is more like a refresher and it doesn't have much proofs mostly problems. Reading on the references page it appears that most of the content is adapted from George Birkhoff and Ralph Beatley's Geometry textbook of the 1940s. – Chris Steinbeck Bell Jun 25 '22 at 00:30
  • https://bookstore.ams.org/chel-120/ – Will Jagy Jun 25 '22 at 01:47