smc

$AHC\sim ACB$ by $AA$, so $\frac{AH}{AC}=\frac{AC}{AH+16}$. Since $AC=15$, we have $\frac{AH}{15}=\frac{15}{AH+16}$. Cross mutliplying, we have $AH(AH+16)=225$. Solving this quadratic yields $AH=9$. Also, $AHC\sim CHB$, so $\frac{AH}{HC}=\frac{HC}{HB}$. Substituting in known values, we have $\frac{9}{HC}=\frac{HC}{16}$, so $H{C}^2=144$ and $HC=12$. The area of $△ABC$ is $\frac{1}{2}(AB)(HC)=\frac{1}{2}(25)(12)=150,\boxed{150}$.