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?

  • $\begingroup$ 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

$ABD$ 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 $\sqrt3$.

  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.


  • $\begingroup$ "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

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 $$ a = 4, \;\; b = 2\sqrt{3}, \;\; c = 2, \;\; d = \sqrt{3}, \;\; m = 1, \;\; n = 3. $$



$ABC$ 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 $BC$.


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