Given $A(0,a),B(-b,0),C(c,0)$. Then $H(0,\frac{bc}{a})$. $AB^{2}=CH^{2} \Rightarrow a^{2}+b^{2}=c^{2}+\frac{b^{2}c^{2}}{a^{2}}$, giving $a=c$ and $\angle ACB=45^{\circ}$.

In ∆BHC, angle HCB=90°-B, angle HBC=90°-C In ∆AHB, angle BAH=90°-B, angle AHB=180°-C In ∆ BHC , By sine rule , sin(90°-B)/BH=sin(90°-C)/HC In ∆AHB, By sine rule, sin(90°-B)/BH= sin(180°-C)/AB So, sin(90°-C)/HC= sin(180°-C)/AB Since , given that CH= AB sin(90°-C)=sin(180°-C) cos C= sin C tan C= 1 So, C= 45° Therefore, angle ACB= 45°

Well known $CH = 2\,OM_a$ (proof in blue, with parallelogramm $AHBA'$) therefore $OM_a = AB/2$ hence $\angle AOB = 90^\circ$ hence $\angle ACB = 45^\circ$ ... or $135^\circ$ !

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My solution: Let $AH$ intersect $BC$ at $S$. Now, we look at $\triangle HCS$ and $\triangle BAS$. We have $\angle HCS = \angle BAS$; $HC = AB$; $\angle HSC = \angle BSA = 90$. Thus, $\triangle HCS = \triangle BAS \implies AS=SC$. But $\angle ASC = 90 \implies \angle ACB = 45$. $\blacksquare$

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