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The problem is as follows:

Find the value of angle $\angle BCA$ or $x$.

Sketch of the problem

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&30^\circ\\ 2.&20^\circ\\ 3.&10^\circ\\ 4.&40^\circ\\ \end{array}$

The thing here is without any further hints. I'm stuck.

I was only able to spot.

$\triangle ABC$ is isosceles,

$\triangle DBC$ is isosceles

But I can't find any other relationship. What else to do?. How to solve this relying only in euclidean geometry postulates?. Can someone help me here?.

Should congruence be used here or what?.

I'm confused please help me. Please include a drawing in your answer as it would help me to spot what to do or if a construction is needed.

Chris Steinbeck Bell
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  • I think you mean $ABD$ is isosceles. But then $AB=BD=BC$, so $ABC$ is isosceles as well, and we are done. – player3236 Jan 10 '21 at 03:59
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    From the problem, this can't be a trapezoid. If $AB||CD$ then $\angle ABD=\angle BDC$. But $70\ne 80$. Similar, for $BC||AD$, you would get $40=50$, which is also wrong. – Andrei Jan 10 '21 at 04:03

5 Answers5

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enter image description here Angle chasing gives $\angle BAD = 50^{\circ}, \angle BCD = 70^{\circ}$. So $\triangle BAD,\triangle BDC$ is isosceles. Hence we have $BA=BD=BC$. Since $BA=BC$, $\triangle BAC$ is isosceles, so $$ x = \frac{180^{\circ}-(80+40)^{\circ}}{2}=30^{\circ}$$

Adola
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As David K points out, we can easily use the left half-triangle and the right half-triangle to show that that $\angle BAD = 50°$ and $\angle BCD = 70°$. So the left and right triangles are isosceles, $BA = BD$, and $BD = BC$, so all three of $A, D, C$ lie on a circle centered at $B$ (not to scale): enter image description here If we extend segment $BC$ out till it reaches the other side of the circle, then $BE$ is a radius since $B$ is the center (although it doesn't look like it), so we actually have diameter $CE$, and arc $EAC$ is actually a semicircle. From here, basic angle-arc definitions and theorems may be used to compute $x$.

Edit: And as another user points out, once we have $BA = BC$, the top half of the diagram is an isosceles triangle with vertex angle of $120$ and base angles of $x$, $x$, which makes the problem very easy!

Rivers McForge
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    In fact, $x+y = 70^\circ,$ so $y = 70^\circ - x$, so $60^\circ - y = x - 10^\circ.$ Also $a+y+70^\circ = 180^\circ,$ so $a = 110^\circ - y = x+40^\circ$ (also obtained by taking $a$ as the exterior angle in a triangle whose interior angles at the other two vertices are $x$ and $40^\circ$). Therefore $180^\circ - a = 140^\circ - x.$ So all angles can be labeled with known angles or angles expressed in terms of $x$. Now what? – David K Jan 10 '21 at 04:09
  • Yeah, I didn't bother to solve it fully out, lol. I figured OP just needed help coming up with variable names. Clearly you're correct that the triangles being isosceles is crucial to solving the problem. I'm going to draw in that circle and add in some comments on that. – Rivers McForge Jan 10 '21 at 04:19
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I think you meant to say $\triangle ABD$ is isosceles. Specifically, $AB = BD.$

You also correctly observe that $\triangle BCD$ is isosceles. Specifically, $BC = BD.$

Putting the equations together, $BA = BC = BD.$ The three points $A, C, D$ are all at the same distance from $B$, so they all lie on a common circle with center at $B.$

That should be a big enough hint.

David K
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enter image description here

Note that after angle chasing we find that $A,C,D$ lies on a circle with center $B$. Note that $$x=70^{\circ}-\angle ACD =70^{\circ} - \frac{1}{2}\angle{ABD}=30^{\circ}$$ by the inscribed angle theorem.

Adola
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enter image description here

(diagram not to scale)

Extend line $AF$ so that it meets the circumference at $F$. As $F,C,D,A$ all lie on the circumference of the circle, quadrilateral $FCDA$ is cyclic.

Hence $\angle BFC = 180º - (50º + 70º) = 60º$ and $\angle FBC = 180º - (80º + 40º) = 60º$, hence $\angle BCF = 60º$. Now since $AF$ is also a diameter, $\angle ACF = 90º$, and thus: $\angle ACB = x = 90º - 60º = \boxed{30º}$.

The same can be done with $E$ instead of $F$, but there is one more step as you would need to find $\angle BAC$ before finding $x$.

Toby Mak
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