Answer:
Final Answer
The length of segment ( BD ) is:
[tex]BD = \frac{5\sqrt{34} - 25}{3}[/tex]
Step-by-step explanation:
Step 1:
[tex]- \( \angle ACB \) \: is \: the \: right \: angle.[/tex]
[tex]- \( AD \) \: is \: the \: angle \: bisector \: of \: \( \angle BAC \), and \: it \: intersects \: \( BC \) \: at \: point \: \( D \).
[/tex]
Step 2:
Using the Pythagorean theorem:
[tex]{AC = \sqrt{AB^2 + BC^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34}}[/tex]
Step 3:
[tex]\frac{BD}{DC} = \frac{AB}{AC}[/tex]
Step 4:
[tex] \frac{x}{3 - x} = \frac{5}{\sqrt{34}}[/tex]
Step 5:
[tex]x \sqrt{34} = 5 (3 - x)[/tex]
[tex]x \sqrt{34} = 15 - 5x[/tex]
[tex]x \sqrt{34} + 5x = 15[/tex]
[tex]x (\sqrt{34} + 5) = 15[/tex]
[tex]x = \frac{15}{\sqrt{34} + 5}[/tex]
Step 6: Rationalize the denominator
[tex]{x = \frac{15}{\sqrt{34} + 5} \times \frac{\sqrt{34} - 5}{\sqrt{34} - 5} = \frac{15(\sqrt{34} - 5)}{34 - 25} = \frac{15(\sqrt{34} - 5)}{9}} [/tex]
[tex]{x = \frac{15\sqrt{34} - 75}{9} = \frac{15}{9} \sqrt{34} - \frac{75}{9} = \frac{5}{3} \sqrt{34} - \frac{25}{3}}[/tex]
