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Problem : In a triangle $ABC$ with side $AB=AC$ and $\measuredangle BAC=20 ^\circ $.

$D $ is a point on side $AC$ and $BC = AD$. Find $ \measuredangle DBC$

Solution:

$AB =AC$

So $ \measuredangle ACB = \measuredangle ABC$

$ \measuredangle ACB = \measuredangle ABC=80^\circ$ (Using angle sum property)

I am unable to continue from here.

Any assistance is appreciated.

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  • $\begingroup$ Are we supposed to do this geometrically or by trig? $\endgroup$ – N.S.JOHN May 12 '16 at 8:41
  • $\begingroup$ any method geometrically or by trig $\endgroup$ – rst May 12 '16 at 8:54
  • $\begingroup$ My calculations lead me to weird stuff. Can x have multiple values? $\endgroup$ – Bradman175 May 12 '16 at 9:40
  • $\begingroup$ NO, multiple values are not there $\endgroup$ – rst May 12 '16 at 9:43
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Let $E$ be a point inside the triangle such that $EB=EC=BC$.

$\qquad\qquad\qquad$enter image description here

Since $\measuredangle{ABE}=\frac{180^\circ-20^\circ}{2}-60^\circ=20^\circ$, we can see that $\triangle{ABD}$ and $\triangle{BAE}$ are congruent.

Hence, $$\measuredangle{DBC}=\measuredangle{ABC}-\measuredangle{ABD}=\measuredangle{ABC}-\measuredangle{BAE}=80^\circ-10^\circ=\color{red}{70^\circ}.$$

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  • $\begingroup$ How did you arrive at this construction? Practice? $\endgroup$ – N.S.JOHN May 12 '16 at 10:49
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    $\begingroup$ @N.S.JOHN: I drew an equilateral triangle $BEC$ in order to "relate" $BC$ to $AD$. ($BC$ moves to $BE$ : $BC$ is "far" from $AD$, but $BE$ is "near" to $AD$ so that we can have triangles with these sides.) $\endgroup$ – mathlove May 12 '16 at 11:14
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Let $ABC$ be the triangle with $O$ being the circumcentre. $AO$ is extended to meet $BC$ at $D$. $BO$ is extended to meet $AC$ at $E$. Let $F$ be the point on $AC$ such that $AF= BC$. Triangle $OAB$ is congruent to triangle $OAC$. Therefore $\angle BAO = \angle CAO = 10^{\circ}$.
Let $BD = a$, so $CD$ also equals $a$ therefore $AB = \frac{a}{\sin 10^{\circ}}$ Now applying sine rule in triangle $AEB$ $$ \frac{AE}{\sin 10^{\circ}} = \frac{AB}{\sin 150^{\circ}} $$ Putting the value of $AB$ as $a/\sin 10^{\circ}$ We get $$ AE = \frac{a}{\sin 30^{\circ}} = 2a. $$ We have $AF= BC = 2a$ Therefore $F$ and $E$ are the same point . Therefore $$ \angle ABF = \angle ABE = 10^{\circ} $$

and hence $$ \angle FBC = 80 - 10 = 70^{\circ}. $$

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