When a particle is projected at an angle $\theta_0$ and again at an angle $(90^{\circ})$ with the same velocity , then although the range R is same in both cases yet the maximum heights are different . If these are $H_1$ and $H_2$ respectively then,

$H_1 =\large\frac{v_0^2 \sin ^2 \theta_0}{2g}$-----(i)

$H_2 = \large\frac{v_0^2 \sin ^2 (90^{\circ}-\theta_0)}{2g}$-----(ii)

Adding (i) and (ii)

$H_1 +H_2 = \large\frac{v_0^2}{2g}$-----(iii)

The maximum height to which a projectile can be projected with velocity $v_0$ when $\theta=45^{\circ}$ is given by,

$H=\large\frac{v_0^2 \sin ^2 45^{\circ}}{2g} = \large\frac{v_0^2}{2(2g)}$------(iv)

From (iii) and (iv)

$H_1+H_2= 2H$