Trig CinarArslan wrote: In a triangle $ABC$ with $\angle ACB = 90^{\circ}$ $D$ is the foot of the altitude of $C$. Let $E$ and $F$ be the reflections of $D$ with respect to $AC$ and $BC$. Let $O_1$ and $O_2$ be the circumcenters of $\triangle {ECB}$ and $\triangle {FCA}$. Show that: $$2O_1O_2=AB$$ Let $M$ be the midpoint of $AC$. Let $\angle CFA = \alpha$. Now, \[\frac{AC}{\sin \alpha} = \frac{CD}{\sin (B-\alpha)} = \frac{BC\sin B}{\sin(B-\alpha)}\implies \cos B = \frac{\sin(B-\alpha)}{\sin \alpha} = \sin B\cot \alpha - \cos B.\]This gives $2\tan \alpha = \tan B$. Therefore, \[\frac{AC}{2MO_2}=\frac{AC}{2BC} \implies MO_2 = BC,\]which gives $O_2B\perp BC$ and $2O_2B = AC$ and we are done.

1-$E,C,F$ linear. 2-Let $M,U,V,K,L$ be midpoints of segments $AB,BC,AC,EC,FC$ respectively. 3-$AE,BF,VK,UL,MC\perp EF$ 4-$O_1=KV\cup MU$, $O_2=LU\cup MV$. 5-$O_1KLU$ perpendicular trapezoid and $MC$ its midsegment so $O_1M=MU$. Similarly $O_2M=MV$. 6-$2O_1O_2=2UV=AB$. Done.

CinarArslan wrote: In a triangle $ABC$ with $\angle ACB = 90^{\circ}$ $D$ is the foot of the altitude of $C$. Let $E$ and $F$ be the reflections of $D$ with respect to $AC$ and $BC$. Let $O_1$ and $O_2$ be the circumcenters of $\triangle {ECB}$ and $\triangle {FCA}$. Show that: $$2O_1O_2=AB$$ Let $M_aM_bM_c$ be the medial triangle of $\triangle ABC$ and let $L, N$ be the midpoints of $EC, FC$ respectively. Claim: $O_1M_bO_2M_a$ forms a parallelogram. Proof: Note that $E, C, F$ are collinear. Also, $L, M_b, O_1$ and $N, M_a, O_2$ are collinear. Also, $O_1M_b \perp EF$ and $O_2M_a \perp EF$. So, $O_1M_b \parallel O_2M_a$. Also, $C$ is the midpoint of $EF$. Thus, $O_1M_c = O_1M_a$ and $O_2M_c = O_2M_b$. So, the claim is true. By our claim, $O_1O_2 = M_aM_b = \frac{AB}{2}$.

Let $O$, $M$, $N$ be midpoint of $AB$, $BC$, $CA$; $G$ be reflection of $C$ through $O$; $O'_1$, $O'_2$ be midpoint of $AG$, $BG$ We have: $\widehat{ECO}$ = $\widehat{ECA}$ + $\widehat{ACO}$ = $\widehat{DCA}$ + $\widehat{CAO}$ = $90^o$ Similarly: $\widehat{FCO}$ = $90^o$ Then: $\overline{E, C, F}$ $\perp$ $AO$ But: $O'_1N$ $\parallel$ $AG$ $\parallel$ $O'_2M$ then: $O'_1N$ $\perp$ $EC$, $O'_2M$ $\perp$ $FC$ So: $O'_1C$ = $O'_1E$, $O'_2C$ = $O'_2F$ But: $OO'_1$ $\parallel$ $CA$ then: $OO'_1$ $\perp$ $BC$ or $O'_1B$ = $O'_1C$ Hence: $O'_1$ is center of ($BCE$) or $O'_1$ $\equiv$ $O_1$ Similarly: $O'_2$ $\equiv$ $O_2$ Therefore: $AB$ = 2 $O_1O_2$

Easy, but cute. Let $T$ be the point such that $ATBC$ is a rectangle. Denote by $R$ and $S$ the intersections of $BF$ with $AD$ and $BD$ with $AE$, respectively. Because $AT$ is the perpendicular bisector of $BS$ and $BT$ is the perpendicular bisector of $AR$, we have that $ASRB$ is a rhombus, thus $SR=AB$, but $O_1O_2$ is precisely the $C$-midline of triangle $CSR$, because $ACFR$ and $ECBS$ are cyclic.