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The problem is indicated in the figure from below:

Sketch of the problem

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&10^{\circ}\\ 2.&15^{\circ}\\ 3.&20^{\circ}\\ 4.&18^{\circ}\\ \end{array}$

What exactly should be done here?. I'm stuck. The only thing which I can spot is that:

$\angle DAC + x = 55$

$\angle BAC + 40 = \angle BCD + 55$

But without any other further knowledge, I don't know what else can be done here?. Can someone help me here?. Can this problem be solved relying only in euclidean geometry?.

The other thing which I can spot is since it says $BD \parallel AC$

This means:

$\angle BDA = \angle DAC$

But again this information doesn't really help me much into solving this problem. Can someone guide me further?. Please include a drawing in the answer because I'm lost. Does this requires a construction?.

Chris Steinbeck Bell
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2 Answers2

8

If $AB=BC=BD$, then $B$ is the centre of circle with $A,C,D$ on the circumference and the angle $\angle ADC$ is half the angle $\angle ABC$, i.e. $20^\circ$

Joffan
  • 39,627
5

enter image description here

At first (blue angles), notice how the $ABC$ triangle is isosceles (since $AB=BC$), and you get that the bottom angles are both $70$ (because $\frac{180-40}{2}=70$). Then, since $BD\| AC$, you can find that the angle next to the $40$º is also $70$, and that the big angle that contains the $x$ is $55$.

Finally (red angles), we also know $AB=BD$, so the triangle $ABD$ is also isosceles and we know one angle is $40+70=110$, so the other two must be $35$ (because $\frac{180-110}{2}=35$).

So your $x$ is $$\boxed{x=55-35=20}$$

Ask me is there's anything you don't understand, maybe I explained it too fast.

Alejandro Bergasa Alonso
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