$\triangle ABC$ is an isosceles triangle such that $AB=AC$ and $\angle BAC$=$20^\circ$. And a point D is on $\overline{AC}$ so that AD=BC, , How to find $\angle{DBC}$?

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I could not get how to use the condition $AD=BC$ , How do I use the condition to find $\angle{DBC}$?

EDIT 1: With MvG's observation, we can prove the following fact.

If we set on a point $O$ in $\triangle{ABC}$ such that $\triangle{OBC}$ is a regular triangle, then $O$ is the circumcenter of $\triangle{BCD}$.

First, we will show if we set a point $E$ on the segment $AC$ such that $OE=OB=OC=BC$, then $D=E$.

Becuase $\triangle{ABC}$ is a isosceles triangle, the point $O$ is on the bisecting line of $\angle{BAC}$. $\angle{OAE}=20^\circ/2=10^\circ$.

And because $OE=OC$, $\angle{OCE}=\angle{OEC}=20^\circ$, $\angle{EOA}=20^\circ-10^\circ=10^\circ=\angle{EAO}$.

Therefore $\triangle{AOE}$ is an isosceles triangle such that $EA=EO$. so $AD=BC=AE$, $D=E$.

Now we can see the point $O$ is a circumcenter of the $\triangle{DBC}$ because $OB=OC=OD.$

By using this fact, we can find $\angle{DBC}=70^\circ$,

  • $\begingroup$ Numerically, $\angle DBC=70°$. A trigonometric expression which evaluates to that shouldn't be to difficult. A proof which doesn't involve the numeric evaluation of trigonometric functions might be a bit more difficult, so if it is that what you need, you might want to make that requirement explicit. $\endgroup$ – MvG Oct 5 '13 at 17:59
  • $\begingroup$ Interesting experimental observation: using $E$ to denote the circumcenter of $BCD$, I found that $BCD$ form a regular triangle, so you have $AD=BC=BE=CE=DE$. Furthermore, $ACDE$ are cocircular, and inscribed angles in that circle on chord $AD$ (or any other of equal length) will be $10°$. Not sure yet how to turn any of this into a proof, but it might help. $\endgroup$ – MvG Oct 5 '13 at 20:17
  • $\begingroup$ Nice Observation, actually we can prove that $\triangle{OBC}$ is a regular triangle such that $O$ is the circumcenter of $\triangle{BCD}$. $\endgroup$ – MS.Kim Oct 6 '13 at 9:10

I saw the following solution may years ago:

On side $AD$ construct in exterior equilateral triangle $ADE$. Connect $BE$.

enter image description here

Then $AB=AC, AE=BC, \angle BAE=\angle ABC$ gives $\Delta BAE =\Delta ABC$ and hence $AB=BE$.

But then $$AB=BE, BD=BD, DA=DE \Rightarrow ADB =EDB$$

Hence $\angle ADB=\angle EDB$. Since the two angles add to $300^\circ$ they are each $150^\circ$. Then $\angle ABD + \angle ADB+ \angle BAD=180^\circ$ gives $ABD=10^\circ$.

  • $\begingroup$ then $\angle{DBC}$ = 30° ( the question is on $\angle{DBC}$ ). Nice ... $\endgroup$ – user354674 Sep 7 '16 at 19:15

One way to calculate this is to write sin laws for two triangles $ABD$ and $BDC$. Call the angle $\angle ABD=x$. Then we have: $$ \frac{AD}{\sin x}=\frac{BD}{\sin 20},\frac{BC}{\sin (20+x)}=\frac{BD}{\sin 80} $$ Using $AD=BC$ and $\sin 80=\cos 10$ we get the following: $$ \frac{\sin x}{\sin 20}=\frac{\sin (20+x)}{\sin 80}\implies \sin x=2{\sin 10}\sin (20+x)\implies\\ \tan x=\frac{2\sin 10\sin 20}{1-2\sin 10\cos 20} $$ Now consider the following identities: $$ 1-2\sin 10\cos 20=1-(\sin 30-\sin 10)=\frac{1}{2}+\sin 10=2\cos 10 \sin 20 $$ Replacing this result in previous equation we get: $$ \tan x=\frac{2\sin 10\sin 20}{2\cos 10\sin 20}=\tan 10 \implies x=10 $$ and hence $\angle DBC=70$.

  • $\begingroup$ Good solution. I was also looking for the ways in which we use only elementary geometry. $\endgroup$ – MS.Kim Oct 6 '13 at 11:10
  • $\begingroup$ @MS.Kim Depending on exactly what you mean by elementary, no such way may exist. Angle ABD is 10 degrees (which trisects 30 degrees), and is not constructible in rigorous compass and straightedge geometry. Angle trisection is one of the three famous impossibilities of Euclidean geometry. $\endgroup$ – David H Oct 6 '13 at 11:33
  • $\begingroup$ @DavidH I understood your comment. I should have said that I was looking for the ways other than trigonometry. Thanks for pointing out it. $\endgroup$ – MS.Kim Oct 6 '13 at 12:07

We construct $\overline{CE}$ such that $E$ lies on $\overline{AB}$ and $\overline{DE}\,||\,\overline{CB}$. We then construct a circle through $A$, $B$ and $C$. Circle $ABC$ has a centre $O$ at the intersection of $\overline{CE}$ and $\overline{BD}$. By the inscribed angle theorem, $\widehat{BOC}=40°$ and since $\triangle BOC$ is isosceles, $\widehat{DBC}=70°$ $\blacksquare$.

  • $\begingroup$ Why does the center of the circle $ABC $ coincide with the intersection point of $BD $ and $AE $? $\endgroup$ – user Apr 2 '20 at 10:19

in $\bigtriangleup$ABC AB=AC and $\angle$A=20. SO $\angle$B=$\angle$C=80

  • NOW WE DRAW $\angle$BAO=60 with AO=AB.AND WE join

    OD and OB

  • now we consider $\bigtriangleup$ABC and $\bigtriangleup$ADO. AD=BC; AO=AB and $\angle$DAO=$\angle$B=80

  • SO $\bigtriangleup$ABC$\cong$$\bigtriangleup$ADO. SO $\angle$AOD=20 and DO=AB=AC

  • furthur we have AO= OB and $\angle$BAO=60.SO $\bigtriangleup$BAO IS equilateral.SO $\angle$AOB =60.FROM THIS we have $\angle$DOB=40.aiso OB=AB=DO

  • OB=DO and $\angle$DOB=40 SO WE HAVE $\angle$OBD=$\angle$ODB=70 and $\angle$OBA=60.SO $\angle$ABD=10

  • $\angle$B=80 so $\angle$DBC=70enter image description here

  • $\begingroup$ Good work for spotting the special relationship between 20, 160, 80, and 60. However, if the vertical angle is x unequal to 20, then (180 - x)/2 = 20 + 60 does not hold. $\endgroup$ – Mick Mar 19 '14 at 19:05

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