Very right quadrilateral

From $C$ drop a perpendicular to $BD$ intersecting at $E$ . By assumption of convexity, $E$ lies on the line segment $BD$ . Observe that right triangles $ABD$ and $DEC$ are congruent by angle-side-angle, hence $CE = BD = 24$ . By Pythagorean theorem on right triangle $BEC$ , we have $BE^2 = BC^2 - CE^2 = 25^2 - 24^2 = 49$ so $BE = 7$ . Thus $AB = ED = BD - BE = 24-7 = 17$ .

Note: If the condition of convexity was not included, then we could have $E$ on the extension of $BD$ , which would give $AB = ED = DB + BE = 24 + 7 = 31$ .