Question 186533
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The Triangle Angle Bisector Theorem is exactly what you need, and gives you plenty of information.


Using the theorem, you can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{l_1}{4}=\frac{l_2}{9}]


Cross-multiplying gives you:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9l_1 = 4l_2 \ \ \Rightarrow\ \ l_1 = \frac{4}{9}l_2]


Now you can substitute in


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l_1^2 + l_2^2 = 13^2]


To get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{4}{9}l_2\right)^2 + l_2^2 = 169]


I'll leave the rest of the arithmetic to you.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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