A geometry problem by Awsaf Rahman

By Pythagorean's Theorem, $AD = \sqrt{21^2 + 72^2} = 75$ , and $CD = \sqrt{75^2 - 45^2} = 60$ .

Let $FG$ be an altitude of $\triangle FBC$ , $EH$ be an altitude of $\triangle EAD$ , and $I$ be the intersection of $AD$ and an extension of $FG$ .

Then $\triangle FBC \sim \triangle GBF$ by angle-angle similarity, so $\frac{FG}{FB} = \frac{FC}{BC}$ or $\frac{FG}{45} = \frac{60}{75}$ , which means $FG = 36$ , and $\frac{BG}{FB} = \frac{FB}{BC}$ or $\frac{BG}{45} = \frac{45}{75}$ , which means $BG = 27$ .

Also, $\triangle EAD \sim \triangle HED$ by angle-angle similarity, so $\frac{HE}{DE} = \frac{AE}{AD}$ or $\frac{HE}{72} = \frac{21}{75}$ , which means $HE = \frac{504}{25}$ , and $\frac{HD}{DE} = \frac{DE}{AD}$ or $\frac{HD}{72} = \frac{72}{75}$ , which means $HD = \frac{1728}{25}$ .

Also, $\triangle IFD \sim \triangle HED$ by angle-angle similarity, so $\frac{IF}{ID} = \frac{HE}{HD}$ . Since $BG = 27$ , then $AI = 27$ , and then $ID = AD - AI = 75 - 27 = 48$ . Therefore, $\frac{IF}{48} = \frac{\frac{504}{25}}{\frac{1728}{25}}$ , which means $IF = 14$ .

The length of $AB$ is the same as $IG$ , and so $AB = IG = IF + FG = 14 + 36 = \boxed{50}$ .