# Geometry : $(AD+DE)^2+BD^2=(AB+BE)^2$

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 $$MK=AM\sin10^{\circ}=\frac{a\sin10^{\circ}}{\sin70^{\circ}}=BD.$$

• I get stuck on one point, to prove that $MK=BD$. I got $BK=BD, \angle BME = \angle BEM = 40^{\circ}, \angle EKD = \angle EDK = 60^{\circ}, \angle KEC = 30^{\circ}, MEK = 110^{\circ}$. I've tried to apply Trigonometry but didn't get the result. Aug 6 '17 at 8:07
• @carat I added something. See now please. Aug 6 '17 at 9:25
• I'm clear now. Thank you for your help. Aug 6 '17 at 10:04
• @carat You are welcome! Aug 6 '17 at 10:24