Let $O$ be the circumcenter of $\triangle ABC$ and $BO,CO$ cut $(O)$ again at $B',C'.$ Since $B',C'$ are the reflections of $H$ about the midpoints of $AC,AB,$ it follows that the composition of homotheties $\mathcal{H}(A, \frac{_1}{^2}) \circ \mathcal{H}(H,2)$ takes $PB,PC$ into $DC',$ $DB'$ $\Longrightarrow$ $\angle C'DB' \cong \angle BPC.$ Let $Q$ be a point on the arc $BC$ not containing $A,$ for which $D$ lies inside $\triangle QB'C'.$ Then $\angle BPC=\angle C'DB' > \angle C'QB'=\frac{_1}{^2}\angle C'OB'=\angle BAC.$

More elementary approach: Since ADPH is parallelogram, DP is parallel to and the same (constant) length, x, as AH. Since AH is an altitude, AH (and therefore DP) is perpendicular to BC. D can be anywhere along BC, but the angle BPC is obviously minimised when D is either at B or at C (easy proof, use calculus if you must). Length x of AH (=DP) is easy trig: $x = b.cosA/sinB = c.cosA/sinC$ which, by sine rule $ = a.cosA/sinA = a/tanA$. The minimum angle of BPC (when D is at B or C) has tangent = a/x, which, by the above formula for x, is simply tan A . Done! Merlin

$\overrightarrow{BP}\cdot \overrightarrow {CP}=BP\cdot CP \cdot \cos{\widehat{BPC}}=AH\cdot BC\cdot \cot{\widehat{BPC}}$. Also, $\overrightarrow{BP}=\overrightarrow{AP}-\overrightarrow{AB}=\overrightarrow{AH}+\overrightarrow{AD}-\overrightarrow{AB}=\overrightarrow{AH}+\overrightarrow{BD}$, and analougously $\overrightarrow{CP}=\overrightarrow{AH}+\overrightarrow{CD}$. Hence, $ AH\cdot BC\cdot \cot{\widehat{BPC}}=\overrightarrow{BP}\cdot \overrightarrow {CP}=(\overrightarrow{AH}+\overrightarrow{BD})(\overrightarrow{AH}+\overrightarrow{CD})=AH^2-BD\cdot CD$.

Observe that $AH^2=AH\cdot 2R\cos{\hat{A}}=AH\cdot BC\cdot \cot{\hat{A}}$, so $AH\cdot BC\cdot \cot{\widehat{BPC}}=AH^2-BD\cdot CD<AH^2=AH\cdot BC\cdot \cot{\hat{A}}$, i.e. $\widehat{BPC}>\widehat{BAC}$ (because the cotangent function is strictly decreasing

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