Here's a problem I just came up with :
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 19⅓ degrees each .
So , can my observation be validated by geometric reasoning or exact calculation ?
Thank you for your feedback . 
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 $E$.
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, $$a=\sqrt{\frac{\sqrt{17}-1}{8}}$$ 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
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 $$ |CM| / |BM| = |AM| / |BM| = |DM| / |AM| = |DM| / |CM|. $$ 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.
We place the triangle on coordinate axes for convenience.
Let $A = (0,0)$, $B = (2b,0)$, $C = (0,2c)$, so that $\angle A = 90^\circ$.
The midpoint of $AC$ is: $$ \left(0, c\right) $$
So the median from $B$ passes through $(2b,0)$ and $(0,c)$.
Using intercept form: $$ \frac{x}{2b} + \frac{y}{c} = 1 $$
Multiplying throughout by $2bc$: $$ cx + 2by = 2bc \quad \text{...(i)} $$
Given $AD \perp BD$, point $D$ is the foot of the perpendicular from $A(0,0)$ to the line (i).
For a line $ax + by + d = 0$, the foot of perpendicular from $(0,0)$ is: $$ \left( \frac{-ad}{a^2 + b^2},; \frac{-bd}{a^2 + b^2} \right) $$
Rewrite (i): $$ cx + 2by - 2bc = 0 $$
So:
Substituting: $$ x = \frac{-c(-2bc)}{c^2 + (2b)^2} = \frac{2bc^2}{c^2 + 4b^2} $$
$$ y = \frac{-(2b)(-2bc)}{c^2 + 4b^2} = \frac{4b^2 c}{c^2 + 4b^2} $$
So, $$ D = \left( \frac{2bc^2}{c^2 + 4b^2},; \frac{4b^2 c}{c^2 + 4b^2} \right) \quad \text{...(ii)} $$
Using the internal angle bisector theorem at $C$:
Points on the angle bisector divide opposite side in the ratio of adjacent sides: $$ \frac{AF}{FB} = \frac{AC}{CB} $$
Now: $$ AC = 2c,\quad CB = \sqrt{(2b)^2 + (2c)^2} = 2\sqrt{b^2 + c^2} $$
So: $$ \frac{AF}{FB} = \frac{2c}{2\sqrt{b^2 + c^2}} = \frac{c}{\sqrt{b^2 + c^2}} $$
Let $F = (x,0)$ on $AB$.
Then: $$ \frac{x}{2b - x} = \frac{c}{\sqrt{b^2 + c^2}} $$
Cross-multiplying: $$ x\sqrt{b^2 + c^2} = c(2b - x) $$
$$ x\sqrt{b^2 + c^2} = 2bc - cx $$
$$ x(\sqrt{b^2 + c^2} + c) = 2bc $$
$$ x = \frac{2bc}{\sqrt{b^2 + c^2} + c} $$
So the x-intercept of the angle bisector is: $$ AF = \frac{2bc}{\sqrt{b^2 + c^2} + c} $$
Thus, equation in intercept form: $$ \frac{x}{\frac{2bc}{\sqrt{b^2 + c^2} + c}} + \frac{y}{2c} = 1 $$
Rewriting: $$ \frac{x(\sqrt{b^2 + c^2} + c)}{2bc} + \frac{y}{2c} = 1 \quad \text{...(iii)} $$
Substitute coordinates from (ii):
First term: $$ \frac{x(\sqrt{b^2 + c^2} + c)}{2bc} = \frac{\frac{2bc^2}{c^2 + 4b^2}(\sqrt{b^2 + c^2} + c)}{2bc} $$
Cancel $2bc$: $$ = \frac{c(\sqrt{b^2 + c^2} + c)}{c^2 + 4b^2} $$
Second term: $$ \frac{y}{2c} = \frac{\frac{4b^2 c}{c^2 + 4b^2}}{2c} = \frac{2b^2}{c^2 + 4b^2} $$
So equation becomes: $$ \frac{c(\sqrt{b^2 + c^2} + c)}{c^2 + 4b^2} + \frac{2b^2}{c^2 + 4b^2} = 1 $$
Combine: $$ \frac{c(\sqrt{b^2 + c^2} + c) + 2b^2}{c^2 + 4b^2} = 1 $$
Multiply both sides: $$ c(\sqrt{b^2 + c^2} + c) + 2b^2 = c^2 + 4b^2 $$
Expand: $$ c\sqrt{b^2 + c^2} + c^2 + 2b^2 = c^2 + 4b^2 $$
Cancel $c^2$: $$ c\sqrt{b^2 + c^2} + 2b^2 = 4b^2 $$
$$ c\sqrt{b^2 + c^2} = 2b^2 $$
Square both sides: $$ c^2(b^2 + c^2) = 4b^4 $$
$$ b^2c^2 + c^4 = 4b^4 $$
Divide by $b^4$, let $\eta = \frac{c}{b}$: $$ \eta^2 + \eta^4 = 4 $$
Rearrange: $$ \eta^4 + \eta^2 - 4 = 0 $$
Let $u = \eta^2$: $$ u^2 + u - 4 = 0 $$
$$ u = \frac{-1 \pm \sqrt{17}}{2} $$
Since $\eta > 0$: $$ \eta = \sqrt{\frac{-1 + \sqrt{17}}{2}} $$
$$ \boxed{ \frac{c}{b} = \sqrt{\frac{\sqrt{17} - 1}{2}} } $$
The rest is easy, coordinates of D,A,B are known, you can check the distances. They are messy, but are equal.
