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Here's a problem I just came up with :

$\triangle ABC$ is a given right triangle at $A$. The median from vertex $B$ and the angle bisector of $C$ intersect at $D$ such that $(AD)$ and $(BD)$ are perpendicular.

I then observed that $\triangle BCD$ is isosceles at $D$!
Indeed, using GeoGebra, I found that the base angles of this triangle measure pratically $19\frac{1}{3}$ degrees each.

So, can my observation be validated by geometric reasoning or exact calculation?

enter image description here

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    $\begingroup$ Is the right angle at $D$ part of the hypotheses, or something you think can be proved? I think you mean it to be part of the hypotheses. Then there in only one such triangle. Please edit the question to clarify - don't reply in a comment. $\endgroup$ Commented Apr 11 at 13:35
  • $\begingroup$ Try using similarity in triangles ADB and ADE ( assuming E is the point where median touches side AC. Then equate the tan of angle ABD and tan of angle ABC . You will get a relation between tanABD and tanABC like : 2 tan ADB = 2 tan ABC I think this will help. $\endgroup$ Commented Apr 11 at 14:14
  • $\begingroup$ I tried to verify, doing some coordinate geometry brute force. the distances you mentioned are indeed equal, you have to set AC=ηBA where η=sqrt[(-1+sqrt(17))/2]. Ya η involves 2 square roots and is very messy. $\endgroup$ Commented Apr 11 at 14:59
  • $\begingroup$ Now you can apply basic SOT and find whatever angles you want :) $\endgroup$ Commented Apr 11 at 15:00
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    $\begingroup$ fyi, as it is isosceles, $\angle DBA=90^\circ-3\circ%$, and using $a$ as given by @ILSG gives $19.334...$, $\endgroup$ Commented Apr 12 at 13:48

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Fun problem! There is no need for computations. Let $M$ be the midpoint of $AC$. One observes that $\triangle ABM$ and $\triangle DAM$ are similar, and therefore $$ \frac{|CM| }{|BM|} =\frac{ |AM| }{ |BM|} =\frac{ |DM| }{|AM|} = \frac{|DM| }{|CM|} $$ As they share angle $\angle CMD = \angle BMC$, it follows that $\triangle CMD$ is similar to $\triangle BMC$. In particular, $\angle CBM = \angle DCM = \angle DCB$ and it follows that $\triangle BCD$ is isosceles.

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    $\begingroup$ A great solution :) $\endgroup$ Commented Apr 12 at 1:40
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    $\begingroup$ Excellent. Thanks. $\endgroup$ Commented Apr 12 at 4:36
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    $\begingroup$ +1 for using just geometry with no trig and no coordinates. $\endgroup$ Commented Apr 12 at 13:03
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We have to do a lot of hunting. Let $\angle ABD=a, \angle DBC=b, \angle ACB= c$.Let the foot of the median be $E$.drawing We see that $$AB\cot c=2AB\tan a\\\implies \tan(a+b)=2\tan a\tag{I}$$This was written due to the fact that $\tan(90°-x)=\cot x$.

By angle bisector theorem, $\frac {CE}{ED}=\frac{CB}{BD}$ Here, $$CE=AE=AB\tan a\\ED=AE\sin a=AB\tan a\sin a\\DB=AB\cos a \\CB=AB\sec (a+b)$$

Expression becomes $$\csc a=\sec (a+b) \sec a\\\implies \sec(a+b)=\cot a\tag{II}$$

Using $\tan^2 x+1=\sec^2 x$, we see $4\tan^2 a+1=\cot^2 a \tag{III}$

For acute angles, $$\tan a=\sqrt{\frac{\sqrt{17}-1}{8}}$$ It gets super easy after this. First, we solve for $b$, using $(I)$. We get $$\tan b=\frac{\tan a}{1+2\tan^2 a}$$ and we know $\tan \frac x2=\frac{\sqrt{1+\tan^2 x}-1}{\tan x}$ for acute $x$, $$\implies \tan \frac c2 =\frac{\sqrt{1+\frac{1}{4\tan^2 a}} -1}{\frac{1}{2\tan a}}\\\implies \tan \frac c2 =\sqrt{1+4\tan^2 a}-2\tan a$$ Using $(III)$, $$\tan \frac c2= \cot a -2 \tan a=\frac{1-2\tan^2 a}{\tan a}\\= \frac{(1-4\tan^4 a)\tan a}{(1+2\tan^2 a)\tan^2 a}=\frac{(\cot^2 a-4\tan^2 a)\tan a}{1+2\tan^ 2 a}=\frac {\tan a }{1+2\tan^ 2 a}=\tan b$$

Hence, it's proven that $\triangle CDB$ is isoceles

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enter image description here

Comment: This particular case takes place when we have an isosceles triangle its sides are $4.1$ units and the attitudes dropped to the sides are exactly $4$ units as shown in the picture.The resulting right angled triangle has those property mentioned in statement that says triangle ADC is isosceles.

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    $\begingroup$ Yes.. Thank you... always the circles....[+1] $\endgroup$ Commented Apr 12 at 11:28
  • $\begingroup$ @JamilSanjakdar, This time the circle is just to show two sides are equal and the triangle is isosceles as a result.+1 for interesting question as always. $\endgroup$ Commented Apr 12 at 13:08
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No trigonometry, no coordinates, no angles ... just some "funny" distances :
Let $M$ the midpoint of [$AC$] , and we can assume that $AC=2$.

$$BA^2 = BD×BM$$ ($\triangle BAD$ and $\triangle BMA$ similar)

$$BA^4 = BD^2(1+AB^2)$$ (Pythagorean Theorem)

$$BD^2 = \frac{BA^4}{(1+AB^2)}$$

On other hand :
$$\frac{DB}{DM} = \frac{BC}{1}$$ (Angle bisector theorem)
$$DB = BC\times DM$$ and $$AM^2 = MD \times MB =1$$ (similar triangles)
$$DB = \frac{BC}{MB} \text{ and } DB^2= \frac{BC^2}{MB^2}$$

$$DB^2= \frac{(AB^2+4)}{(AB^2+1)} = \frac{AB^4}{(1+AB^2)} $$

Finally, $$4DM^2=2(DC^2+DA^2)- AC^2$$ (median theorem)
with $$AD^2=AB^2-BD^2 , DM^2=1-AD^2 \text{ and } BD^2=\frac{AB^4}{(1+AB^2)} $$

Thus , we find : $$DC^2=\frac{(4+AB^2)}{(1+AB^2)}=BD^2$$ and $\triangle DBC$ isosceles.

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