Let $A_1, B_1, C_1$ and $A_2, B_2, C_2$ be the corners of its bases. Let the radius of the sphere be $r$, so $h = 2r$. View from above is drawn below ($T_1$ and $T_2$ are directly on top of each other, so they are shown as a single point $T$, also note that $O_1T_1$ and $O_2T_2$ are equal to $O_1T$ and $O_2T$) [asy][asy] size(8cm); defaultpen(fontsize(10pt)); import olympiad; pair A = dir(120), B = dir(210), C = dir(330), T = (-0.3, -0.1); pair D = foot(T, B, C), E = foot(T, A, C), F = foot(T, A, B); real radius = 0.3; path ct = circle(T, radius); pair xa = tangent(D, T, radius, 1), ya = tangent(D, T, radius, 2); pair xb = tangent(E, T, radius, 1), yb = tangent(E, T, radius, 2); pair xc = tangent(F, T, radius, 1), yc = tangent(F, T, radius, 2); pair AA = extension(xb, yb, xc, yc), BB = extension(xa, ya, xc, yc), CC = extension(xa, ya, xb, yb); pair O = circumcenter(A, B, C); pair OO = circumcenter(AA, BB, CC); draw(A -- B -- C -- cycle); draw(AA -- BB -- CC -- cycle); draw(ct); draw(T -- D ^^ T -- E ^^ T -- F); draw(A -- AA, dashed); draw(B -- BB, dashed); draw(C -- CC, dashed); dot("$A_1$", A, dir(120)); dot("$B_1$", B, dir(210)); dot("$C_1$", C, dir(330)); dot("$A_2$", AA, dir(140)); dot("$B_2$", BB, dir(160)); dot("$C_2$", CC, dir(20)); dot("$T$", T, dir(100)); dot("$O_1$", O, dir(70)); dot("$O_2$", OO, dir(-50)); [/asy][/asy] Let the angles of the similar triangles be $\alpha$, $\beta$, $\theta$. Let the distances of point $T$ to sides of the two bases be $x_1, y_1, z_1, x_2, y_2, z_2$. Observation: $x_1\cdot x_2 = y_1 \cdot y_2 = z_1 \cdot z_2 = r^2$ Proof: The plane $A_1A_2B_2B_1$ is tangent to the sphere. It is easy to see this observation from the side view. [asy][asy] size(6cm); defaultpen(fontsize(10pt)); import olympiad; pair A = (0, 0), B = (9, 0), C = (0, 12), D = (4, 12), E = (0, 6); pair F = foot(E, B, D); draw(A -- B, L = Label("$x_1$", position = MidPoint)); draw(D -- C, L = Label("$x_2$", position = MidPoint)); draw(B -- D ^^ A -- C); draw(E -- F, L = Label("$r$", position = MidPoint)); draw(circle(E, 6)); [/asy][/asy] Lemma: $T$ is a point inside of a triangle with angles $\alpha$, $\beta$, $\theta$ and distances from its sides are $x, y, z$. Then the equation below holds, where $O$ is the circumcenter of the triangle and $R$ is its radius.$$\frac{R^2-OT^2}{2R}=\frac{yz\sin{\alpha}+xz\sin{\beta}+xy\sin{\theta}}{x\sin{\alpha}+y\sin{\beta}+z\sin{\theta}}$$Proof: ...Trigonometry, I will post this part later... [asy][asy] size(8cm); defaultpen(fontsize(10pt)); import olympiad; pair A = dir(120), B = dir(210), C = dir(330), T = (-0.3, -0.1); pair O = circumcenter(A, B, C); pair D = foot(T, B, C), E = foot(T, A, C), F = foot(T, A, B); draw(A -- B -- C -- cycle); dot("$T$", T, dir(90)); dot("$O$", O, dir(90)); draw(T -- D, L = Label("$x$", position = MidPoint)); draw(T -- E, L = Label("$y$", position = MidPoint)); draw(T -- F, L = Label("$z$", position = MidPoint)); [/asy][/asy] Applying the lemma for $(x, y, z) = (x_1, y_1, z_1)$ and for $(x, y, z) = (x_2, y_2, z_2) = (\frac{r^2}{x_1}, \frac{r^2}{y_1}, \frac{r^2}{z_1})$, we get these: $$\frac{R_1^2-O_1T_1^2}{2R_1} = \frac{y_1z_1\sin{\alpha}+x_1z_1\sin{\beta}+x_1y_1\sin{\theta}}{x_1\sin{\alpha}+y_1\sin{\beta}+z_1\sin{\theta}}$$$$\frac{R_2^2-O_2T_2^2}{2R_2} = \frac{\frac{r^4}{x_1y_1z_1}(x_1\sin{\alpha}+y_1\sin{\beta}+z_1\sin{\theta})}{\frac{r^2}{x_1y_1z_1}(y_1z_1\sin{\alpha}+x_1z_1\sin{\beta}+x_1y_1\sin{\theta})}$$Multiplying these two equations gives the desired result.