0
$\begingroup$

Consider $\triangle ABC$ with a point $D \in BC$. If $\triangle ABD \sim \triangle ADC$ in ratio $\frac {1}{\sqrt3}$. Find the angles of $ABC$

My try

Well, the first thing i saw is if $ABD$ and $ADC$ are similar, then if i'm not wrong, $AD$ is the angle bisector of $ABC$, because $\angle$DAC =$\angle$BAD so i played with the similarity ratio and put $x$ and $y$ to some lengths.

My Measurements

Ex:$\vartriangle$$ABD$ $\sim$ $\vartriangle$$ADC$

then $$\frac{AB}{AD}=\frac{BD}{DC}=\frac{AD}{AC}=\frac{1}{\sqrt3}$$

Let $AB=x$

$$\frac{x}{AD}=\frac{1}{\sqrt3}$$

$$x\sqrt3=AD$$

I did this with all the sides of the triangles, then i applied Angle bisector theorem with the measures and i ended with $\sqrt 3=3$ (obviously a wrong equality).

Sorry for not putting all my calculations here, but i think it doesn't matter because they are wrong.

I think my mistake is in the use of the ratios. Are my approaches good and i just put bad the ratios, or i missed something? any hints?

No trigonometry allowed

$\endgroup$
  • $\begingroup$ The way you write similiarity you have $AB||AD$??? $\endgroup$ – Aqua Feb 15 '18 at 6:19
  • $\begingroup$ No, just because the triangles are similar the lengths of corresponding sides are proportional $\endgroup$ – Rodrigo Pizarro Feb 15 '18 at 6:22
0
$\begingroup$

We should notice that there is a unique triangle that satisfies these conditions. For example if we try to take $\angle ABC > 90^\circ$, then we cannot have $\triangle ABD \sim \triangle ADC$ because $\angle ADC = \angle ABD + \angle BAD$.

If we try to take $\triangle ABC$ acute, then if $AD$ is not altitude of $\triangle ABC$, then either $\angle ADC > 90^\circ$ or $\angle ADB > 90^\circ$. For both cases, one of $\triangle ABD$ and $\triangle ADC$ is acute while the other one is wide-angle triangle. If $AD$ is altitude, for acute $\triangle ABC$, we can satisfy $\triangle ABD \sim \triangle ADC$ with $\angle BAD = \angle DAC$ but clearly, this gives us the ratio $1$, which is not we want.

Therefore, the triangle we seek is a special one where $\angle BAC = 90^\circ$. After that, if we place the angles as the following in order to have a ratio different than $1$, we can see that $\triangle ABC$ is $30^\circ-60^\circ-90^\circ$ triangle (WLOG, we can take $|AD| = 1$ and find the corresponding lengths with respect to $|AD|$).

enter image description here

$\endgroup$

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.