Splitting $AB^2+AC^2=BC^2$

First we observe that $\triangle ABC \sim \triangle ABD$ because their angles are congruent. Then, we can find the following relationship:

$\displaystyle{\frac{BD}{AB}=\frac{AD}{AC}}$

To find the length of $AD$ we use the Pythagorean Theorem:

$BD^2+AD^2=AB^2$

$3^2+AD^2=24^2$

$9+AD^2+576$

$AD=\sqrt{567}=\sqrt{3^4\cdot7} =9\sqrt{7}$

Now we can find the length of $AC$ :

$\frac{BD}{AB}=\frac{AD}{AC}$

$\frac{3}{24}=\frac{9\sqrt{7}}{AC}$

$3\cdot AC=24\cdot 9\sqrt{7}$

$AC=72\sqrt{7}$

Since this answer is in its reduced form and the problem asks for $a+b$ from a value in the form $a\sqrt{b}$ , the answer you should input is $72+7=\boxed{79}$

The famous Pythagorean Theorem tells us $AB^2 + AC^2 = BC^2$ .

So, we can say─

$AB^2+AC^2 = BC \cdot BC$

$\implies AB^2+AC^2 = BC \cdot BC$

$\implies AB^2+AC^2 = BC \cdot (BD+CD)$

$\implies AB^2+AC^2 = BC \cdot BD + BC \cdot CD$

Now let's split-up the last equality. Do the following equalities hold? $AB^2 = BC \cdot BD$ $AC^2 = BC \cdot CD$

Yes, they hold (when $AD \perp BC$ ).

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• From the similarity between $\Delta ABC$ and $\Delta DBA$ , we have

$\frac{AB}{BD} = \frac{BC}{AB}$ , which proves $AB^2 = BC \cdot BD$

• Similarly, one can prove . $AC^2 = BC \cdot CD$ from the similarity between $\Delta ABC$ and $\Delta DCA$ ,

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Now, As $AB^2 = BC \cdot BD$ , we get $BC = \frac{AB^2}{BD} = \frac{24^2}{3} = 192$ .

So, $AC = \sqrt{BC^2-AB^2} = \sqrt{192^2-24^2} = \sqrt{36288} = \sqrt{2^6 \cdot 3^4 \cdot 7} = 2^3 \cdot 3^2 \sqrt{7} = 72\sqrt{7} = a\sqrt{b}.$

Therefore, $a= 72, b = 7$ with $a+b = \boxed{79}$ .