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The problem is as follows:

Sketch of the problem

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\textrm{4 m}\\ 2.&\textrm{5 m}\\ 3.&\textrm{6 m}\\ 4.&\textrm{7 m}\\ \end{array}$

How exactly should this be solved relying only in euclidean geometry?.

I'm confused exactly if this should use congruence or what?. The part which I'm stuck is what to do with that the double of the angle on $\angle ACD$?.

I'm stuck can someone help me here?. Please include a drawing in your answer because I'm not very savvy with understanding where exactly congruence could be

Semiclassical
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Chris Steinbeck Bell
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  • In case anyone misses it: The drawing has angle $\alpha=\angle BCD$ whereas $2\alpha=\angle ACD$. (It's a bit obscured by the TV mast.) – Semiclassical Jan 11 '21 at 02:59

2 Answers2

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Triangle ABC

Reflect $\triangle ABD$ over segment $AB$ so that point $D$ ends up in a new position, point $E$.

Observe that ,in $\triangle AED$ and $\triangle CAE$, $\angle EAD=2\alpha=\angle ACE$ and both the triangle share the same angle $\angle E$. Hence, they are similar and thereafter $\triangle CAE$ is isosceles as well.

Thus, $CE=10$ and $CD=CE-DE=10-2\cdot 2=10-4=\boxed {6}$

Limestone
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    Howdy!, I'm still stuck. Why $\triangle AED$ and $\triangle CAE$ are similar?. You mention it's because they share the same angle but I've looked upon the cases of similarity on triangles and these are SAS, SSS and AA. Which of these applies here?. Can you help me?. Can you clear this doubt please?. – Chris Steinbeck Bell Jan 12 '21 at 19:17
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    If you know one angle of an isosceles triangle, you know the other two... – Semiclassical Jan 12 '21 at 20:52
  • @Chris Steinbeck Bell Since we have already deduced that $\angle EAD=\angle CAE=2\alpha$ and both the triangles share the same angle $\angle E$, it can be said that the triangles are similar. Hence if one of them is isosceles, the other one has to be isosceles as well with the respective equivalent sides being equal. – Limestone Jan 13 '21 at 03:04
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    @Chric Steinbeck Bell When two triangles have two of their angles equal to two other angles of the other triangle ,they are similar. – Limestone Jan 13 '21 at 03:55
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    @DRSKMOBINULHAQUE Thanks for the explanation, I do understand now. The source of confusion was to spot where were the angles. It seems that those two are referring to $\angle E$ and $EAD$. – Chris Steinbeck Bell Jan 13 '21 at 23:30
  • @Chris Steinbeck Bell Glad that you liked it. – Limestone Jan 14 '21 at 04:06
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A trigonometric solution is straightforward. We have:

$$\frac{2}{AB} = \tan \alpha = \frac {\sin \alpha}{\cos \alpha}$$

$$\frac {AB}{10} = \sin 2\alpha = 2\sin \alpha \cos\alpha$$

Multiplying the two equations, we have:

$$\frac2{10} = 2\sin^2\alpha$$

which gives $\sin \alpha = \dfrac {1}{\sqrt{10}}$. This gives $AD = 2\sqrt{10}$, and by applications of the Pythagorean Theorem we can find the lengths of $AB$, $BC$ and finally $DC$.

player3236
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