Let $ABC$ be a right triangle where $\angle C=90^{\circ}$ and $\angle A=10^{\circ}$.
Point D and E are on the sides AC and BC respectively such that $\angle ABD = \angle CDE = 60^{\circ}$.
Prove that $(AD+DE)^2+BD^2=(AB+BE)^2$
My attempt :
Draw line $BD$ and extend $BD$ through $D$ to meet the perpendicular from $A$ at $F$, $AF\perp BD$
Let $ED$ cut $AF$ at point $T$.
$\angle CBF = \angle CAF = 20^{\circ} \rightarrow B, C, F, A$ concyclic.
so $\angle BCA = \angle BFC = \angle BDE = 10^{\circ}$
so $DE \parallel CF$ and $ET \parallel CF$.
Let $M\in AB$ such that $B$ placed between $M$ and $A$ and $MB=BE$ and
$K\in AC$ such that $C$ be a midpoint of $KD$.
Hence, since $\measuredangle CDE=60^{\circ}$, we have $AK=AD+DK=AD+DE$.
Also, $AM=AB+BM=AB+BE$.
Thus, it remains to prove that $MK=BD$ and $\measuredangle MKA=90^{\circ}$.
From here I used trigonometry:
We can show that a projection of $MB$ on the line $AC$ is equale to $KC$,
which gives that $\measuredangle MKA=90^{\circ}$.
Now, we can show that $MK=BD$ and it ends the proof by the Pythagoras theorem.
Maybe it will help and you'll find something nicer.
I'll prove that $MK=BD$ if you proved that $MK\perp AK$.
Let $AB=a$.
Hence, by law of sines for $\Delta ABD$ we obtain: $$\frac{BD}{\sin10^{\circ}}=\frac{a}{\sin70^{\circ}}$$ or $$BD=\frac{a\sin10^{\circ}}{\sin70^{\circ}}.$$ Now, by law of sines for $\Delta BED$ we obtain: $$\frac{BE}{\sin10^{\circ}}=\frac{\frac{a\sin10^{\circ}}{\sin70^{\circ}}}{\sin150^{\circ}},$$ which gives $$MB=BE=\frac{2a\sin^210^{\circ}}{\sin70^{\circ}}.$$ Thus, $$AM=AB+BM=a+\frac{2a\sin^210^{\circ}}{\sin70^{\circ}}=\frac{a(\sin70^{\circ}+1-\cos20^{\circ})}{\sin70^{\circ}}=\frac{a}{\sin70^{\circ}},$$ which says $$MK=AM\sin10^{\circ}=\frac{a\sin10^{\circ}}{\sin70^{\circ}}=BD.$$