No angle allowed!

Take triangle $ADC$

$AC^{2} = AD^{2} + DC^{2}$

$5^{2} = 4^{2} + DC ^{2}$

$25 - 16 = DC^{2}$

$3 = DC$

Take $DB$ as $x$ .

In triangle $ABC$ ;

$5^{2} +BC^{2} = (4 + x)^{2}$

$16 + x^{2} + 8x -25 = BC^{2}$ $(1)$

In triangle $BCD$

$DC^{2} + BD^{2} = BC^{2}$

$3^{2} + x^{2} = BC^{2}$ $(2)$

Equating 1 & 2

$16 + x^{2} + 8x -25 = BC^{2} = 3^{2} + x^{2} = BC^{2}$

$8x = 18$

$x = 2.25$

Therefore; $AB = 4 + 2.25 = 6.25$

By Pythagoras Theorem; 6.25^{2} - 5^{2) = BC^2

$39.0625 - 25 = BC^{2}$

$14.0625 = BC^{2}$

$BC = 3.75$

We could determine that CD = 3 through Pythagorean Triplets.

Then if we draw an arc with radius equal to 5, we could determine that BC is a tangent and is equal to $5 \tan A$ . So BC is:

$BC = 5 \tan A = 5\Big(\frac{3}{4}\Big)$ $\boxed{BC = 3.75}$

$\angle BAC=\angle CAD$ and $\angle ACB=\angle ADC=90°$ leaving $\angle ABC=\angle ACD$ .

This makes ABC and ACD similar triangles so we can just compare ratios.

CD is 3/4th of AD, since it is 3 (per the most famous example of Pythagorean Theorem) and $3=0.75*4$ .

BC is 3/4th of AC, making it 3.75 because $3.75=0.75*5$ .