Drag vertex A to explore. The triangle is constrained to remain acute.
Consider an acute-angled triangle $ABC$. $D$ is the orthocenter of $\triangle ABC$. Denote the circumcircle of $\triangle ABC$ and the circumcircle of $\triangle DBC$ by $\Gamma_1$ and $\Gamma_2$ respectively. $AD$ produced intersects $BC$ and $\Gamma_1$ at points $H$ and $J$ respectively.
We need to determine the validity of:
Click on the "Proof" tabs above to explore step-by-step why all three statements must be true.
$\angle BDC + \angle BAC = 180^\circ$
Let $AE$ and $AF$ be the altitudes from $A$. Since $D$ is the orthocenter, $\angle AED = \angle AFD = 90^\circ$.
Consider the quadrilateral AFDE. The sum of its interior angles is $360^\circ$.
Therefore, $\angle FDE + \angle FAE = 360^\circ - 90^\circ - 90^\circ = 180^\circ$.
Notice that $\angle BDC$ and $\angle FDE$ are vertically opposite angles, so $\angle BDC = \angle FDE$.
Conclusion: Substituting the angles, we get $\angle BDC + \angle BAC = 180^\circ$. Statement I is True.
$DH = HJ$
Let's compare right-angled $\triangle BDH$ and $\triangle JBH$:
Therefore, $\angle JBC$ = $\angle EBC$ (which is $\angle DBH$). Both are $90^\circ - \angle C$.
Since $\triangle BDH$ and $\triangle JBH$ share side $BH$, $\angle DBH = \angle JBH$, and $\angle BHD = \angle BHJ = 90^\circ$, we have $\triangle DBH \cong \triangle JBH$ (A.S.A.).
Thus, corresponding sides are equal: $DH = HJ$. Statement II is True.
Radius of $\Gamma_1$ = Radius of $\Gamma_2$
Because $DH = HJ$ and $BC \perp DJ$, $J$ is the geometric reflection of $D$ across the line $BC$.
Therefore, the circumcircle of $\triangle DBC$ ($\Gamma_2$) is a direct reflection of $\Gamma_1$ across $BC$, ensuring identical radii! Statement III is True.
(Don't forget that there is one and only one circle passing through 3 non-collinear points.)