Given $\triangle ABC$ with orthocenter $H$. $BH=\frac{c \cdot \cos \beta}{\sin \gamma}=2R \cos \beta$ and $CH=\frac{b \cdot \cos \gamma}{\sin \beta}=2R \cos \gamma$, radius $R$ of the circumcircle. $BH+CH=2R(\cos \beta+\cos \gamma)=4r=4(p-a)\tan \frac{\alpha}{2}$ $R(\cos \beta+\cos \gamma)=2(p-a)\tan \frac{\alpha}{2}$ $R(2\cos \frac{\beta+\gamma}{2}\cos \frac{\beta-\gamma}{2})=2(p-a)\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}$. $R(2\cos \frac{\beta-\gamma}{2})=2(p-a)\frac{1}{\cos \frac{\alpha}{2}}$. $R(2\cos \frac{\alpha}{2}\cos \frac{\beta-\gamma}{2})=2(\frac{a+b+c}{2}-a)$. $R(2\sin \frac{\beta+\gamma}{2}\cos \frac{\beta-\gamma}{2})=b+c-a$. $R(\sin \beta+\sin \gamma)=b+c-a$. $2R(\sin \beta+\sin \gamma)=2(b+c)-2a$. $b+c=2(b+c)-2a$ and $b+c=2a$, perimeter $=a+b+c=3a$.