4
$\begingroup$

Triangle ABC has point D on BC which creates triangle ABD and ACD. They differ with the scale factor $\sqrt{3}$. What are the angles?

I know ADB and ADC cannot be right, as it shares the side AD and cannot have a scale factor of $\sqrt{3}$.

I have tried to approach it with $\frac{AD}{DC}=\frac{BD}{AD}=\sqrt{3}$ (or $\frac{AD}{AC}=\frac{BA}{AD}=\sqrt{3}$). Meaning that $DC=x$ $AD=\sqrt{3}x$ $BD=3x$. This is where I get stuck. Because I dont know if the triangle is right angled, and cannot use Pythagoras. I dont have any angles, hence cannot use $\frac{sin(a)}{angleA}$. How do I solve this?

$\endgroup$
  • $\begingroup$ Yes, exactly. Changed it! $\endgroup$ – Oscar3 Jul 19 '17 at 20:42
  • $\begingroup$ Aha, you meant to put "searching for angles" in the title? $\endgroup$ – Oscar3 Jul 19 '17 at 20:44
  • $\begingroup$ @WW1 But how do I find the third side of either ABD or ACD? $\endgroup$ – Oscar3 Jul 19 '17 at 20:47
  • $\begingroup$ I disagree with your argument for the triangles not being right. I'm pretty sure that's a possibility. $\endgroup$ – Shuri2060 Jul 19 '17 at 20:47
  • $\begingroup$ @Shuri2060 Im not saying that they are not, but how do I know that they are? Can I somehow prove it? $\endgroup$ – Oscar3 Jul 19 '17 at 20:48
1
$\begingroup$

$ABD$ and $ACD$ are similar and differ by scale factor $\sqrt3$.

We therefore know that $\angle ADB=\angle ACD\lor\angle ADB=\angle CAD\lor\angle ADB=\angle ADC$.

Let us consider the first case, where $\angle ADB=\angle ACD$.

Then $AD \,||\, AC$. But this is impossible unless $C$ and $D$ coincide which is also impossible due to the scale factor.

Let us consider the second case, where $\angle ADB=\angle CAD$.

Then $\angle ABD\neq\angle BAD$ as ${AD\over AD}\neq\sqrt3$. Therefore $\angle ABD=\angle ADC$ leading to a similar contradiction as before.

Hence we conclude $\angle ADB=\angle ADC$.

$B,C,D$ are collinear $\implies\angle ADB=\angle ADC = \frac{\pi}{2}$.

$\angle BAD\neq\angle CAD$ as ${AD\over AD}\neq\sqrt3$. Therefore $\angle BAD=\angle ACD$ and $\angle ABD=\angle CAD$.

wlog, let $ABD$ be the larger triangle and let $CD = x$.


You should now be able to fill out all of the lengths using the ratio of lengths of the similar triangles and Pythagoras (also, notice $\angle BAC = {\pi\over2}$).

You'll end up seeing you have $1 : \sqrt3 : 2$ right-angled triangles which you should know the angles of.


Finally, you may want to convince yourself that such a construction is possible. The way the question is set up implies that there is a solution, so there is no need to check once you find the only possibility.

However, outside of such a context, this may simply be done by first constructing $ACD$ and then letting $B$ be a point collinear to $CD$ such that $\angle BAC ={\pi\over2}$. Next, show that the triangles are similar and have a scale factor $\sqrt3$.

$\endgroup$
2
$\begingroup$
  1. Let $\Delta ABD\sim\Delta ACD$.

Hence, $\frac{AB}{AC}=\frac{AD}{AD}$, which is impossible.

  1. Let $\Delta ABD\sim\Delta ADC$.

Hence, $\measuredangle ABD=\measuredangle ADC,$ which is impossible.

  1. Let $\Delta ABD\sim\Delta CAD$.

Hence, $\measuredangle ADB=\measuredangle CDA=90^{\circ}$ and since $\frac{AD}{CD}=\sqrt3$, we have $\measuredangle C=60^{\circ}$

and from here $\measuredangle B=30^{\circ}$ and $\measuredangle BAC=90^{\circ}$.

  1. Let $\Delta ABD\sim\Delta CDA$.

Hence, $\measuredangle ABD=\measuredangle CDA$, which is impossible.

  1. Let $\Delta ABD\sim\Delta DAC$.

Hence, $\measuredangle BAD=\measuredangle ADC$, which says $AB||BC$, which is impossible.

  1. Let $\Delta ABD\sim\Delta DCA$, which gives $AB||BC$ again.

Done!

$\endgroup$
  • $\begingroup$ "Hence, $∡ADB=∡CDA=90∘∡ADB=∡CDA=90$" How do you know that? $\endgroup$ – Oscar3 Jul 20 '17 at 8:49
  • $\begingroup$ @Oscar3 If $\Delta ABD=\Delta CAD$ then $\measuredangle ADB=\measuredangle CDA$. But $\measuredangle ADB+\measuredangle CDA=180^{\circ}$, which says $\measuredangle ADB=\measuredangle CDA=90^{\circ}$ $\endgroup$ – Michael Rozenberg Jul 20 '17 at 8:53
  • $\begingroup$ @Oscar3 In my previous comment I meant $\Delta ABD\sim\Delta CAD$. It was typo. $\endgroup$ – Michael Rozenberg Jul 20 '17 at 9:43
1
$\begingroup$

A solution is given by dropping an altitude to the hypotenuse of a 30-60-90 right triangle. Is this the only one? To find out we define the six quantities $a = BC$, $b = CA$, $c = AB$, $d = AD$, $m = BD$, and $n = DC$. There are four ways in which the triangles $ABD$ and $ACD$ could be similar: \begin{align} 1)& \;\; \frac{b}{c} = \frac{d}{m} = \frac{n}{d} = \sqrt{3},\\ 2)& \;\; \frac{b}{m} = \frac{d}{c} = \frac{n}{d} = \sqrt{3},\\ 3)& \;\; \frac{b}{d} = \frac{d}{c} = \frac{n}{m} = \sqrt{3},\\ 4)& \;\; \frac{b}{d} = \frac{d}{m} = \frac{n}{c} = \sqrt{3}.\\ \end{align} (The other two permutations have $d/d = \sqrt{3}$, so we can rule them out immediately.) We need to solve for six quantities. Each of the cases above reduces this to three. The relationship $a = m+n$ reduces it two. Setting one of the lengths to 1 reduces it one. So we only need one more relationship. For this we use Stewart's Theorem: $$ man + dad = bmb + cnc. $$ Cases 1 through 3 each have 1 solution with positive side lengths, and case 4 has 2 solutions. However, only in case 1 do the three resulting triangles all satisfy the triangle inequality. This solution is the 30-60-90 right triangle $$ a = 4, \;\; b = 2\sqrt{3}, \;\; c = 2, \;\; d = \sqrt{3}, \;\; m = 1, \;\; n = 3. $$

$\endgroup$
0
$\begingroup$

Hint:

$ABC$ is a rectangular triangle with $\hat A=90°$ and $\hat B=60°$ .

Take $D$ such that $AD$ is the height of the triangle with respect to $BC$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.