Question

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as shown in the figure, point d is a point on $overline{ab}$, a side of $\triangle abc$. $ad = 2, db = 3$, and if $mangle b=mangle acd, ac=$.

Explanation:

Step1: Prove triangle similarity

Since $\angle B=\angle ACD$ and $\angle A=\angle A$ (common - angle), by the AA (angle - angle) similarity criterion, $\triangle ACD\sim\triangle ABC$.

Step2: Use the property of similar - triangles

For similar triangles $\triangle ACD$ and $\triangle ABC$, the ratios of corresponding sides are equal. That is, $\frac{AC}{AB}=\frac{AD}{AC}$.
We know that $AB = AD + DB=2 + 3=5$ and $AD = 2$.
Substituting the values into the proportion $\frac{AC}{5}=\frac{2}{AC}$, we get $AC^{2}=AD\times AB$.

Step3: Solve for $AC$

$AC^{2}=2\times5 = 10$. Then $AC=\sqrt{10}$.

Answer:

$\sqrt{10}$