1

The problem is as follows:

In the figure, $AC=2AB$. Using this information find $\alpha$.

Sketch of the problem

The choices given in my book are:

$\begin{array}{cc} 1.&30^{\circ}\\ 2.&37^{\circ}\\ 3.&\frac{45}{2}^{\circ}\\ 4.&\frac{53}{2}^{\circ}\\ 5.&35^{\circ}\\ \end{array}$

According to the official answers sheet the answer for this is choice 1. But how to get to that answer?.

You could use trigonometry and with some effort get to the answer. But this is not the solution I am looking for.

What I don't know is how to relying only in euclidean geometry do some construction or anything to solve this problem. Can it be done?.

If you look carefuly you may conclude that it might be something to do with a 30-60-90 right triangle because one side is two times of the other the same as their angles. But how to prove using constructions that one of those vertex is a right angle?

Since, I am not good with constructions and it seems this requires doing one. Therefore, can someone help me here?.

Chris Steinbeck Bell
  • 4,175

1 Answers1

1

Pick a point $P$ on $BC$ such that the triangle $APC$ is an isoceles triangle with angles $\alpha$ on either side, then pick a point $Q$ on $AC$ such that $\Delta APC$ is split into two right angled triangles $\Delta APQ$ and $\Delta BPQ$, where $Q$ is the midpoint of $AC$. From the conditions, $AQ=QC=AB$.

$\Delta APQ$ has a side $AP$ in common with $APB$, and $AQ=AB$. (I've marked this as $a$ in the diagram.) Since the angle between the two equal sides is $\alpha$ in both triangles, $\Delta ABP$ and $\Delta APQ$ are congruent, hence $\angle ABP$ is a right angle.

From the sum of the angles in $\Delta ABC=90^\circ+3\alpha=180^\circ$ we get $\alpha=30^\circ$.

enter image description here

Suzu Hirose
  • 11,660
  • Did you use any devices to make and pst the picture? My friend has been teaching from home and uses some computer thing he refers to as a scribbler: he and the student can see each others digrams, calculations... – Will Jagy Jun 20 '22 at 23:15
  • I used paint dot net to make that, but there are better things you can use I think, I just have no idea how to use those. – Suzu Hirose Jun 20 '22 at 23:16
  • Thank you, I will look into that. So far, I draw something on paper, put it on a scanner to make a jpeg. – Will Jagy Jun 20 '22 at 23:22
  • @SuzuHirose Thanks for that. Well explained, just as I was looking for. I'd like to note that drawing $PQ$ is the altitude of $\triangle{APC}$ and being isosceles makes it a perpendicular bisector and the criteria used is $\triangle{APQ}\cong\triangle{ABP}$ by means of the case SAS. Please confirm if I understood well. – Chris Steinbeck Bell Jun 22 '22 at 02:04
  • @WillJagy By the way, if you are interested. I use Inkscape for making these drawings and its quite good. – Chris Steinbeck Bell Jun 22 '22 at 02:07
  • @SuzuHirose Did you read my words from above?. Is what I mentioned good?. Please I appreciate you attend this doubt. – Chris Steinbeck Bell Jun 24 '22 at 23:55
  • @ChrisSteinbeckBell please don't ping me like this. – Suzu Hirose Jun 25 '22 at 00:31