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The alternatives given in my book are as follows:

Sketch of the problem

$\begin{array}{ll} 1.&30^{\circ}\\ 2.&40^{\circ}\\ 3.&20^{\circ}\\ 4.&45^{\circ}\\ \end{array}$

I'm stuck with this problem. What exactly should it be done here?. The thing is that I was only able to spot these relationships:

$\angle ABD = \angle ADB$

I'm assuming the intended approach is to draw a line between $DC$. But I can't really find a case of congruence to find the requested angle. Can someone help me here?.

How to solve this using only euclidean geometry?. The thing here is that. I'm stuck. I guess that the approach maybe is to spot an isosceles?. But other than the ones stated there I can't find others.

I've attempted to look if this figure could be inscribed in a circle but it doesn't seem the case. So I don't know what else to do?.

Can someone help me here please?. Please include a drawing in the answer so I can understand what to do.

Chris Steinbeck Bell
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    I think you've typo'd or read the question wrong. Do you mean $\angle BAD = \angle ADB$? And yes joining $BC$ (not $DC$, but I think you mean $BC$) is the right direction. – Tony Ip Jan 07 '21 at 00:45
  • @TonyIp Sorry I made a mistake in typing what I meant was what you mentioned. However this does not explain why $BC=BD$? Where did you got that relationship? – Chris Steinbeck Bell Jan 07 '21 at 01:40
  • Because if you have a triangle with two equal sides (AB=AC in ABC) in which at least one angle is $60^\circ$, it's the equilateral triangle, and BD equals any of its sides by the initial assumption. – Thehx Jan 07 '21 at 03:09

3 Answers3

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enter image description here The inspiration here is to join $BC$. Note that $\triangle ABC$ is equilateral (why?), hence we have $BD=BC=AB$. That means we can draw a circle with $B$ as the center and $BA, BC, BD$ as the radii of the circle.

Can you work from here?

Tony Ip
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  • Assuming that you intend to follow the strategy of using the chords in the inscribed triangle this would meant that the mysterious angle is $30^{\circ}$ because is half of what is the subtended arc with the center on $B$. But how did you concluded $BC=BD$? Please explain because this part I'm lost. – Chris Steinbeck Bell Jan 07 '21 at 01:42
  • Hmm... are you unsure on why $\triangle ABC$ is equilateral? Because if you're okay with this claim, we'll have $AB=BC$, and together with the given condition that $AB=BD$, we'll have the desired relation. – Tony Ip Jan 07 '21 at 01:47
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The triangle ABC is equilateral from the givens and, in turn, A, C and D are cyclic with their circumcenter B. Thus, $x$ = $\frac12\angle$ABC = $\frac12\cdot 60^\circ = 30^\circ$.

Quanto
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enter image description here

$$AB = AC = BD$$

$$m\angle BAD = 50^\circ$$

$$m\angle DAC = 10^\circ$$

Since $m\angle BAC = 60^\circ$ and $AB = AC$, then $\triangle ABC$ is an equilateral triangle.

So $AB = BC = CA$.

So $m\angle ABC = m\angle ACB = 60^\circ$.

So $m\angle AFB = m\angle ACF + m\angle FAC = 70^\circ$. (External angle theorem.)

Since $AB = BD$, $\triangle ABD$ is isosceles. So $m\angle ADB = 50^\circ$ and $m\angle ABD = 80^\circ$.

Since $BC = BD \dots$

Steven Alexis Gregory
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