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In the following problem, I want to find the angle marked as $x$. It seems so simple and yet I am out of ideas. It is very easy to get all angles except two of them: angle ADB and angle CBD.

Is there a calculation for the angle $x$ that only uses parallel lines and no circles?

Edit: A nice solution by using a circle is given below. In the book, this question was asked at the start of the chapter after introducing parallel lines and all the angles in that setup. Sum of angles of a triangle was proved and this question was asked in the exercise. 1

From geogebra, I know the answer is 60 degrees. But I do not know how to argue that the answer is 60 degrees.

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  • $\begingroup$ I solved your problem. If you want to see my solution, show please your attempts. $\endgroup$ – Michael Rozenberg Jun 8 at 8:23
  • $\begingroup$ Hello @MichaelRozenberg, I do not know what to show in failed attempts of drawing parallel lines, chasing angles and reaching dead ends. There is a key construction that I am missing. If it was a routine calculation, I would have definitely shown my attempts. $\endgroup$ – Isomorphism Jun 8 at 10:38
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Given:
1) $\angle ABC=30^\circ$
2) $\angle BAD=80^\circ$
3) $\angle DAC=20^\circ$
4) $\angle ACB=50^\circ$
5) $\angle BCD=50^\circ$


1) (the key point of this solution) Let $F\in AB$ such, that $\angle BCF$ $=\angle FBC=30^\circ$, $\angle ACF$ $=\angle CAD=20^\circ$
2) $\angle ADC $ $= 180^\circ-\angle DAC-\angle ACB-\angle BCD$ $=180^\circ-20^\circ-50^\circ-50^\circ$ $=60^\circ$
3) $\angle AFC $ $= \angle FCB+\angle FBC$ $=30^\circ+30^\circ$ $=60^\circ=\angle ADC$ $\Rightarrow$
4) $ACDF$ is cyclic $\Rightarrow$
5) $\angle CFD=\angle CAD=20^\circ$
6) $\angle CDF$ $=180^\circ-\angle CFD-\angle FCB-\angle BCD$ $=180^\circ-20^\circ-30^\circ-50^\circ$ $=80^\circ$ $\Rightarrow$
7) $\angle FCD=\angle FDC$ $\Rightarrow$
8) $DF=CF$
9) $CF=FB$ $\Leftarrow$ $\angle FBC=\angle FCB=30^\circ$
10) (8-9$\Rightarrow$)$DF=FB$ $\Rightarrow$ $\angle FDB$ $=\angle FBD$ $=\frac{180^\circ-\angle BFD}{2}$ $=\frac{180^\circ-(\angle BFC-\angle CFD)}{2}$ $=\frac{\angle BCF+\angle FBC+\angle CFD}{2}$ $=\frac{30^\circ+30^\circ+20^\circ}{2}=40^\circ$
11) $\angle ADF=\angle ACF$ $\Leftarrow$ (4)
12) $\angle ADB=\angle ADF+\angle FDB=20^\circ+40^\circ=60^\circ$
13) Bonus: $\angle DFB=\angle CAB$ $\Rightarrow$ $AC||DF$

I can't see any parallel lines or circles as well.

So, construct them)
Seriously, $\angle BCF=30^\circ$ is the first thing that it wants to construct.
If needed with no circles, $\triangle AEF$ and $\triangle CED$ are congruent, thus $\triangle DEF$ is equilateral and similar to $\triangle CEA$, but I think it's a bit longer.

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  • $\begingroup$ All the answers are nice, but I will accept this answer since it has a parallel line. I could start with the parallel line construction and solve the problem. $\endgroup$ – Isomorphism Jun 9 at 23:56
  • $\begingroup$ In fact this approach would be hard so I didn't mention it. Constructing such $F$ and then deriving $AC||DF$ is quite easy, but when you construct $AC||DF,\,F\in AB$ then proving $\angle FCB=30^\circ$ requires non-obvious steps, you can try it. $\endgroup$ – Alexey Burdin Jun 10 at 0:03
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    $\begingroup$ I felt that the parallel line approach was easier.. Noting CDFA is an isosceles trapezoid gives triangles CDA and AFC congruent, and also triangle ADF is congruent to CFD. We then get DF = CF and CF=FB. Now DFB is isosceles with F=100, so we are done. $\endgroup$ – Isomorphism Jun 10 at 5:08
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Let $AD\cap BC=\{E\}$.

Thus, since $$\frac{AE}{CE}\cdot\frac{CE}{DE}\cdot\frac{DE}{BE}\cdot\frac{BE}{AE}=1$$ we obtain: $$\frac{\sin50^{\circ}}{\sin20^{\circ}}\cdot\frac{\sin60^{\circ}}{\sin50^{\circ}}\cdot\frac{\sin(70^{\circ}-x)}{\sin{x}}\cdot\frac{\sin80^{\circ}}{\sin30^{\circ}}=1$$ or $$\frac{\sin(70^{\circ}-x)}{\sin{x}}=\frac{\sin20^{\circ}\sin30^{\circ}}{\sin60^{\circ}\sin80^{\circ}}.$$ But it's obvious that $x$ is an acute angle and $\frac{\sin(70^{\circ}-x)}{\sin{x}}$ decreases.

Id est, it's enough to prove that: $$\frac{\sin10^{\circ}}{\sin60^{\circ}}=\frac{\sin20^{\circ}\sin30^{\circ}}{\sin60^{\circ}\sin80^{\circ}}$$ or $$2\sin10^{\circ}\sin80^{\circ}=\sin20^{\circ},$$ which is obvious.

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    $\begingroup$ Thank you for a wonderful solution. Sorry that I could not produce a meaningful attempt. $\endgroup$ – Isomorphism Jun 8 at 15:03
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We draw first the red rhombus $ABFE$ with angles $100^\circ$ and $80^\circ$. Then we draw the lines $g$, $h$ through $A$ and $F$ which make an angle of $20^\circ$ with two sides of the rhombus. Due to symmetry these lines intersect at a point $D$ on the symmetry axis $BE$ of the rhombus. All black noted angle sizes in the figure can now be immediately verified. It is easily seen that all angles at $D$, in particular the angle $x$, are $60^\circ$.

Since $|CF|=|EF|$ the triangle CFB is isosceles. This certifies the $50^\circ$ of the median angle at $C$, and in a second step the $50^\circ$ of the lower angle there, as well as the angle $30^\circ$ at $B$. It follows that our quadrilateral $ABDC$ is a copy of your $ABDC$.

enter image description here

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  • $\begingroup$ Beautiful! Thank you for sharing. $\endgroup$ – Isomorphism Jun 10 at 5:09
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Let $O$ be the circumcenter of $\triangle ABC$. Then since $\angle COA = 2\angle CBA=60^\circ$ and $OA=OC$, triangle $COA$ is equilateral. Easy angle chasing reveals that $\angle CDA=60^\circ = \angle COA$ and therefore $CDOA$ is cyclic. It follows that $\angle DOC=\angle DAC=20^\circ$. On the other hand, since $O$ is circumcenter of $\triangle ABC$ and $\angle BAC=100^\circ$, we see that the convex angle $COB$ is $360^\circ-2\angle BAC = 160^\circ$. It follows that $D, O, B$ are collinear. Now, since $COA$ is equilateral, $AC=OA$, and since $CDOA$ is cyclic, $DA$ bisects the angle $CDO$. Therefore $$x=\frac 12 \angle CDO =\frac 12 \cdot 120^\circ = 60^\circ.$$

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  • $\begingroup$ Thank you for your solution. This question was asked before the introduction of circles. Only parallel lines and angles in a transversal configuration was introduced before this problem. I would be interested to see a proof with just parallel lines as well. $\endgroup$ – Isomorphism Jun 8 at 10:31
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after erecting equilateral triangle

From angles in $\triangle ABC$ we have that $\angle CBA=30^\circ$. From angles in $\triangle CAD$ we have that $\angle CDA=60^\circ$.

Erect an equilateral $\triangle$ $BCG$ on base $BC$. Then $\angle GCB=60^\circ$ so $\angle GCA=10^\circ$, and $\angle CBG=60^\circ$ so $\angle ABG=30^\circ$. Join $DG$. From angles in $\triangle CGD$, $\angle DGC=10^\circ$. $\angle DAC=20^\circ$ so $DG$ is parallel to $DA$ produced, and so $A$ is on $DG$. (In more detail: Let $CA$ (extended if necessary) meet $DG$ at $A′$. Then $\angle CA′G=160^\circ$ so $\angle DA′C=20^\circ=\angle DAC$, so $A′=A$.)

$\angle CBG=60^\circ=\angle CDG$, so quadrilateral $CGBD$ is cyclic. Therefore $\angle ADB=\angle GDB=\angle GCB=60^\circ$.

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