I have been given the figure below:
I know that $AD=20-x$ and $m\angle ACD=m\angle BCD$. How can I set up a ratio also knowing that $AC=11$ and $BC=14$ in order to find $x$? Thanks!
2026-03-25 12:51:56.1774443116
Solving for length of an unknown side of a triangle.
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We can write: $$\dfrac{S_{\Delta ACD}}{S_{\Delta CDB}}=\dfrac{AD}{BD}$$ And Also: $$\dfrac{S_{\Delta ACD}}{S_{\Delta CDB}}=\dfrac{CD\cdot CA\cdot\sin(\angle ACD)}{CD\cdot CB\cdot \sin(\angle BCD)}=\dfrac{CA}{CB}$$ So we can conclude: $$\dfrac{AD}{BD}=\dfrac{CA}{CB}\Rightarrow\dfrac{20-x}{x}=\dfrac{11}{14}$$
And so on.