I also had trouble with this question.
16. In ABC with a right angle at C, point D lies in the interior of AB and point E lies in the interior of BC so that AC=CD, DE=EB, and the ratio AC:DE = 4:3. What is the ratio AD:DB?
A) 2:3
B) 2: \(\sqrt5\)
C) 1:1
D) 3:\(\sqrt5\)
E) 3:2
16.
In ABC with a right angle at C,
point D lies in the interior of AB and
point E lies in the interior of BC so that
AC=CD, DE=EB, and the ratio AC:DE = 4:3.
What is the ratio AD:DB?
\(\text{Let $\overline{AC} = \overline{CD} = {\color{red}x} $ } \\ \text{Let $\overline{DE} = \overline{EB} = {\color{red}y} $ } \\ \text{Let $\overline{AD} = {\color{red}s} $ } \\ \text{Let $\overline{DB} = {\color{red}t} $ } \)
\(\text{Let $\angle{CAD} = \angle{ADC} = \alpha $ } \\ \text{Let $\angle{ABC} = \angle{EDB} = 90^{\circ}-\alpha $ } \\ \text{Let $\angle{DCA} = 180^{\circ}-2\alpha $ } \\ \text{Let $\angle{ECD} = 90^{\circ}-\angle{DCA} = 90^{\circ} -(180^{\circ}-2\alpha) = 2\alpha-90^{\circ} $ } \\ \text{Let $\angle{BED} = 180^{\circ}-2(90^{\circ}-\alpha)=2\alpha $ } \\ \text{Let $\angle{DEC} = 180^{\circ}-\angle{BED}=180^{\circ}-2\alpha $ } \)
\(\begin{array}{|rcll|} \hline \angle{CDE} &=& 180^{\circ}- \angle{ECD}-\angle{DEC} \\ &=& 180^{\circ}-(2\alpha-90^{\circ})- (180^{\circ}-2\alpha ) \\ &\mathbf{=}& \mathbf{90^{\circ} \ !} \\ \hline \end{array} \)
\(\mathbf{\text{$\cos$-rule in $\triangle ACD$}} \)
\(\begin{array}{|rcll|} \hline x^2 &=& x^2+s^2-2xs\cos(\alpha) \\ 2xs\cos(\alpha) &=& s^2 \\ \mathbf{2x\cos(\alpha)} & \mathbf{=} & \mathbf{s} \quad & (1) \\ \hline \end{array}\)
\(\mathbf{\text{$\cos$-rule in $\triangle DEB$}}\)
\(\begin{array}{|rcll|} \hline y^2 &=& y^2+t^2-2yt\cos(90^{\circ}-\alpha) \\ 2yt\sin(\alpha) &=& t^2 \\ \mathbf{2y\sin(\alpha)} & \mathbf{=} & \mathbf{t} \quad & (2) \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline \dfrac{(1)}{(2)}: & \dfrac{ 2x\cos(\alpha) } {2y\sin(\alpha) } &=& \dfrac{ s }{t} \\ \\ & \dfrac{x}{y} \cdot \dfrac{ \cos(\alpha) } {\sin(\alpha) } &=& \dfrac{ s }{t} \\ \\ & \mathbf{ \dfrac{x}{y} \cdot \dfrac{ 1 } {\tan(\alpha) }} & \mathbf{=} & \mathbf{\dfrac{ s }{t}} \quad & (3) \\ \hline \end{array}\)
\(\mathbf{\text{In the right-angled $\triangle DCE$}} \\ \mathbf{ \tan(\alpha) =\ ?}\)
\(\begin{array}{|rcll|} \hline \tan(180^{\circ}-2\alpha) &=& \dfrac{x}{y} \\\\ -\tan(2\alpha) &=& \dfrac{4}{3} \quad | \quad \text{Formula:} \ \boxed{ \tan(2\alpha)=\dfrac{2\tan{\alpha}}{1-\tan^2(\alpha)} } \\\\ \dfrac{-2\tan{\alpha}}{1-\tan^2(\alpha)} &=& \dfrac{4}{3} \\\\ \dfrac{2\tan{\alpha}}{\tan^2(\alpha)-1} &=& \dfrac{4}{3} \\\\ \dfrac{ \tan{\alpha}}{\tan^2(\alpha)-1} &=& \dfrac{2}{3} \\\\ 3\tan{\alpha} &=& 2(\tan^2(\alpha)-1) \\ 3\tan{\alpha} &=& 2\tan^2(\alpha)-2 \\ 2\tan^2(\alpha)-3\tan(\alpha)-2 &=& 0 \\\\ \tan(\alpha)&=& \dfrac{3\pm \sqrt{9-2\cdot4\cdot(-2)}} {2\cdot 2} \\\\ \tan(\alpha)&=& \dfrac{3\pm \sqrt{25}} {4} \\\\ \tan(\alpha)&=& \dfrac{3\pm 5} {4} \\\\ \tan(\alpha)&=& \dfrac{3 {\color{red}+} 5} {4} \quad | \quad \tan(\alpha) > 0\ ! \\\\ \tan(\alpha)&=& \dfrac{8} {4} \\ \mathbf{ \tan(\alpha) }& \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{ \dfrac{x}{y} \cdot \dfrac{ 1 } {\tan(\alpha) }} & \mathbf{=} & \mathbf{\dfrac{ s }{t}} \quad | \quad \mathbf{\tan(\alpha)=2},\ \quad \dfrac{x}{y}=\dfrac{4}{3} \\\\ \dfrac{4}{3} \cdot \dfrac{1} {2} & = & \dfrac{ s }{t} \\\\ \dfrac{4}{6} & = & \dfrac{ s }{t} \\\\ \dfrac{2}{3} & = & \dfrac{ s }{t} \\\\ \mathbf{\dfrac{ s }{t}} &\mathbf{=}& \mathbf{\dfrac{2}{3}} \\ \hline \end{array} \)
\((A) \ 2:3\)
16.
In ABC with a right angle at C,
point D lies in the interior of AB and
point E lies in the interior of BC so that
AC=CD, DE=EB, and the ratio AC:DE = 4:3.
What is the ratio AD:DB?
\(\text{Let $\overline{AC} = \overline{CD} = {\color{red}x} $ } \\ \text{Let $\overline{DE} = \overline{EB} = {\color{red}y} $ } \\ \text{Let $\overline{AD} = {\color{red}s} $ } \\ \text{Let $\overline{DB} = {\color{red}t} $ } \)
\(\text{Let $\angle{CAD} = \angle{ADC} = \alpha $ } \\ \text{Let $\angle{ABC} = \angle{EDB} = 90^{\circ}-\alpha $ } \\ \text{Let $\angle{DCA} = 180^{\circ}-2\alpha $ } \\ \text{Let $\angle{ECD} = 90^{\circ}-\angle{DCA} = 90^{\circ} -(180^{\circ}-2\alpha) = 2\alpha-90^{\circ} $ } \\ \text{Let $\angle{BED} = 180^{\circ}-2(90^{\circ}-\alpha)=2\alpha $ } \\ \text{Let $\angle{DEC} = 180^{\circ}-\angle{BED}=180^{\circ}-2\alpha $ } \)
\(\begin{array}{|rcll|} \hline \angle{CDE} &=& 180^{\circ}- \angle{ECD}-\angle{DEC} \\ &=& 180^{\circ}-(2\alpha-90^{\circ})- (180^{\circ}-2\alpha ) \\ &\mathbf{=}& \mathbf{90^{\circ} \ !} \\ \hline \end{array} \)
\(\mathbf{\text{$\cos$-rule in $\triangle ACD$}} \)
\(\begin{array}{|rcll|} \hline x^2 &=& x^2+s^2-2xs\cos(\alpha) \\ 2xs\cos(\alpha) &=& s^2 \\ \mathbf{2x\cos(\alpha)} & \mathbf{=} & \mathbf{s} \quad & (1) \\ \hline \end{array}\)
\(\mathbf{\text{$\cos$-rule in $\triangle DEB$}}\)
\(\begin{array}{|rcll|} \hline y^2 &=& y^2+t^2-2yt\cos(90^{\circ}-\alpha) \\ 2yt\sin(\alpha) &=& t^2 \\ \mathbf{2y\sin(\alpha)} & \mathbf{=} & \mathbf{t} \quad & (2) \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline \dfrac{(1)}{(2)}: & \dfrac{ 2x\cos(\alpha) } {2y\sin(\alpha) } &=& \dfrac{ s }{t} \\ \\ & \dfrac{x}{y} \cdot \dfrac{ \cos(\alpha) } {\sin(\alpha) } &=& \dfrac{ s }{t} \\ \\ & \mathbf{ \dfrac{x}{y} \cdot \dfrac{ 1 } {\tan(\alpha) }} & \mathbf{=} & \mathbf{\dfrac{ s }{t}} \quad & (3) \\ \hline \end{array}\)
\(\mathbf{\text{In the right-angled $\triangle DCE$}} \\ \mathbf{ \tan(\alpha) =\ ?}\)
\(\begin{array}{|rcll|} \hline \tan(180^{\circ}-2\alpha) &=& \dfrac{x}{y} \\\\ -\tan(2\alpha) &=& \dfrac{4}{3} \quad | \quad \text{Formula:} \ \boxed{ \tan(2\alpha)=\dfrac{2\tan{\alpha}}{1-\tan^2(\alpha)} } \\\\ \dfrac{-2\tan{\alpha}}{1-\tan^2(\alpha)} &=& \dfrac{4}{3} \\\\ \dfrac{2\tan{\alpha}}{\tan^2(\alpha)-1} &=& \dfrac{4}{3} \\\\ \dfrac{ \tan{\alpha}}{\tan^2(\alpha)-1} &=& \dfrac{2}{3} \\\\ 3\tan{\alpha} &=& 2(\tan^2(\alpha)-1) \\ 3\tan{\alpha} &=& 2\tan^2(\alpha)-2 \\ 2\tan^2(\alpha)-3\tan(\alpha)-2 &=& 0 \\\\ \tan(\alpha)&=& \dfrac{3\pm \sqrt{9-2\cdot4\cdot(-2)}} {2\cdot 2} \\\\ \tan(\alpha)&=& \dfrac{3\pm \sqrt{25}} {4} \\\\ \tan(\alpha)&=& \dfrac{3\pm 5} {4} \\\\ \tan(\alpha)&=& \dfrac{3 {\color{red}+} 5} {4} \quad | \quad \tan(\alpha) > 0\ ! \\\\ \tan(\alpha)&=& \dfrac{8} {4} \\ \mathbf{ \tan(\alpha) }& \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{ \dfrac{x}{y} \cdot \dfrac{ 1 } {\tan(\alpha) }} & \mathbf{=} & \mathbf{\dfrac{ s }{t}} \quad | \quad \mathbf{\tan(\alpha)=2},\ \quad \dfrac{x}{y}=\dfrac{4}{3} \\\\ \dfrac{4}{3} \cdot \dfrac{1} {2} & = & \dfrac{ s }{t} \\\\ \dfrac{4}{6} & = & \dfrac{ s }{t} \\\\ \dfrac{2}{3} & = & \dfrac{ s }{t} \\\\ \mathbf{\dfrac{ s }{t}} &\mathbf{=}& \mathbf{\dfrac{2}{3}} \\ \hline \end{array} \)
\((A) \ 2:3\)
Mathcourts24,
How about saying thankyou to heueka for all his time consuming effort which he afforded you!
Don'r just grumble about getting it wrong!
You tried to vote your question up about 10 times but you have not even attempted to give Heureka a point.
That is incredibly rude of you! Do you think that will encourage anyone to help you next time?