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The problem is as follows:

Using the figure from below: Find the unknown angle indicated as $x$. Assume $AD=BC$ and $BD=DC$

Sketch of the problem

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&12^{\circ}\\ 2.&10^{\circ}\\ 3.&15^{\circ}\\ 4.&16^{\circ}\\ 4.&14^{\circ}\\ \end{array}$

What I attempted to do here was to add the angles in the isosceles which adds up to $4x$, this can be added to the $3x$ triangle but that's where I'm stuck.

In other words the only thing which I could spot was:

$\angle BDA = 2x+2x$

How exactly can it be used congruency to solve this problem?. Can someone help me here?. The intended approach is relying in euclidean geometry postulates, but I don't know exactly which sort of congruency of triangles identity should be used.

Please include a drawing in your answer because this part is difficult for me to spot with accuracy. Can you please use an explanation step-by-step.

Chris Steinbeck Bell
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2 Answers2

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Draw a line $DF$ such that $\angle ADF = 3x$. Draw lines $AE$ and $DE$ such that $\angle DAE = \angle ADE = 2x$.

enter image description here

So $\triangle AED \cong \triangle BDC$ (by A-A-S, as $BC = AD$)

So, $AE = DE = BD = DC$

Now $AF = DF$ so $\triangle AFE \cong \triangle DFE$ (by S-A-S).

Similarly, $\triangle DFB \cong \triangle DFE$

So, we have $\angle AFE = \angle EFD = \angle DFB = 60^0$

So $\angle FAD = \angle FDA = 3x = 30^0$

So, $x = 10^0$.

Math Lover
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  • I'm wondering if such complicated construction was that necessary to solve this problem. In your solution you mention $\triangle AED \cong \triangle BDC$ but in the figure it doesn't seem to be that way. By the way did you used geogebra or any other similar software to draw your answer?. – Chris Steinbeck Bell Nov 22 '20 at 21:07
  • This specific problem is easier solved by just applying sine law but your tag was Euclidean geometry. I mention $\triangle AED \sim \triangle BDC$ because I take angles of $2x$. So it has to be. In the drawing I did not measure and draw. Yes I think to prove it without trigonometry, such a construct is necessary. Also I do not think it is that complicated because all I am doing is making isosceles triangles based on $2x$ nd $3x$ angles that are given. So I can get to $x$. – Math Lover Nov 22 '20 at 21:14
  • I'm sorry but I cannot distinguish very well why is it A-A-S in $\triangle AED$ and $\triangle BDC$ and S-A-S in the triangles $\triangle AFE$ and $\triangle DFE$ can you explain which are the angle angle and side you are referring to and the side angle side you were also referring to the second triangle. I'm trying to follow your explanation but I'm stuck with them, please help me with this part. – Chris Steinbeck Bell Nov 23 '20 at 08:39
  • In the first case we are drawing angles of $2x$ from points $A$ and $D$ so $\angle ADB = \angle BDA = \angle DBC = \angle DCB = 2x$. Also the question says $BC = AD$. Is the first case clear why A-A-S? Once you acknowledge, will go to the next. – Math Lover Nov 23 '20 at 08:52
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enter image description here

In the figure we have $$\dfrac{a}{\sin(7x)}=\dfrac{b}{\sin(3x)}\\\dfrac{a}{\sin(4x)}=\dfrac{b}{\sin(2x)}$$ It follows $$\sin(7x)=2\sin(3x)\cos(2x)$$ You have two options: solve this last equation or check which of the five answers given is correct. By both means (the second is preferable) you find $x = 10^{\circ}$(in boths sides you have $0.93969262$).

►If you have difficulty to solve the equation note that $$2\sin(3x)\cos(2x)=\sin(5x)+\sin(x)$$ and$$\sin(7x)-\sin(5x)=2\cos(6x)\sin(x)$$ Thus $$2\cos(6x)\sin(x)=\sin(x)\Rightarrow \cos(6x)=\dfrac12\Rightarrow x=10^{\circ}$$

Piquito
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