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The Figure of Problem

Where $\angle{A}=84^{\circ}, \angle{ACD}=42^{\circ}, BD=AC$, find $\angle{BCD}$.

Wonder if there is solution without using trigonometric functions.

I tried with getting circumcenter of triangle ABC, but seems hard to prove it forms an equilateral triangle with side AC.

Also if trying from equilateral and form an Isosceles triangle with two angles of $24^{\circ}$, and then form another isosceles triangle with top angle to be $\angle{B}=24^{\circ}$, it is not easy to prove that $\angle{ADC}=54^{\circ}$.

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    $\begingroup$ Please tell what you have tried and where you are stuck $\endgroup$
    – user987987
    Nov 3, 2021 at 8:14
  • $\begingroup$ I tried with getting circumcenter of triangle ABC, but seems hard to prove it forms an equilateral triangle with side AC. $\endgroup$
    – r ne
    Nov 3, 2021 at 8:50
  • $\begingroup$ There is only one relationship: $\angle{B}+\angle{BCD}=54^{\circ}$. This problem is not solvable by angle chasing, at least not this way. $\endgroup$
    – r ne
    Nov 3, 2021 at 9:11
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    $\begingroup$ @ACB Added texts. $\endgroup$
    – r ne
    Nov 3, 2021 at 10:31

2 Answers 2

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Draw isosceles triangle $ACE$ such that $CA=CE$.

step1

Let the circumcentre of $\triangle DEC$ be $F$ and draw the equilateral triangle $DEF$.

step2

$\angle CEF=\angle ECF=36^\circ$.

Now construct another equilateral triangle with side $CE$ as in the figure.

step3

$\angle DEG=36^\circ$

$\therefore\triangle DEG\cong\triangle FEC\text{ (S-A-S)}\\ \implies DE=DG.$

Also $\angle BDG=72^\circ.$

Then mark point $H$ on $CE$ such that $CF=CH$.

step4

$\therefore \angle CHF=\angle CFH=72^\circ$.

Also $\triangle EHF$ is isosceles.

Now we can see $\triangle EDG\sim\triangle EHF$. From there we can prove that $\triangle BGE\sim \triangle CFE$.

From similar isosceles triangles above, $\frac{ED}{EG}=\frac{EH}{EF}.$
Adding $1$ to both sides, $\frac{EB}{EG}=\frac{EC}{EF}\implies\small \triangle BGE\sim \triangle CFE.$

It shows that $BG=GE$ and then we can easily show $\angle EBC=\angle ABC=30^\circ.$

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    $\begingroup$ (+1) Very nice! How did you come up with the idea of adding equilateral triangles and the first isosceles triangle, if I may ask? $\endgroup$
    – Saeed
    Nov 16, 2021 at 5:55
  • $\begingroup$ “Let the circumcentre of △DEC be F and draw the equilateral triangle DEF.” But how do we know that △DEF is equilateral, not merely isosceles like △ECF and △DCF? $\endgroup$ Dec 4, 2021 at 14:10
  • $\begingroup$ The solution is great. Thanks and sorry for my late response! $\endgroup$
    – r ne
    Dec 6, 2021 at 22:52
  • $\begingroup$ Got a simpler solution, starting with the same idea but ending with shorter approach: i.imgur.com/SmmC6B2.png $\endgroup$
    – r ne
    Dec 7, 2021 at 0:41
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    $\begingroup$ @TimPederick , $\angle DFE=2\angle DCE$ (why?) $\endgroup$
    – ACB
    Dec 12, 2021 at 11:56
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enter image description here

Working "the other way around" sometimes helps. Consider triangle $\triangle ACD$, and construct the regular pentagon $PQRDC$ as shown above, noting that $AD$ bisects $\angle CDR$. Finally produce $AD$ to $B^\prime$ in such a way that $\triangle CRB^\prime$ is equilateral.

We will show that $B \equiv B^\prime$, by proving that $B^\prime D \cong AC$, as in the triangle in OP's hypotheses.

  1. Angle chasing yields $\measuredangle ACP = \measuredangle CAP = 76^\circ$. Thus $\measuredangle CPA = 48^\circ$.
  2. Working similarly on $\triangle ARQ$ allows to claim that $\triangle PAQ$ is equilateral.
  3. Angle chasing and $CQ \cong RB^\prime$ imply that $\triangle B^\prime DR \cong \triangle ACQ$ (SAS), and, in particular $B^\prime D \cong AC$.

So, as stated, $B^\prime\equiv B$, and, in particular $$\boxed{\measuredangle DCB = 24^\circ}.$$

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