# In a triangle $ABC$ with side $AB=AC$ and $\measuredangle BAC=20 ^\circ$. $D$ is a point on side $AC$ and $BC = AD$. Find $\measuredangle DBC$

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.

• Are we supposed to do this geometrically or by trig? – N.S.JOHN May 12 '16 at 8:41
• any method geometrically or by trig – rst May 12 '16 at 8:54
• My calculations lead me to weird stuff. Can x have multiple values? – Bradman175 May 12 '16 at 9:40
• NO, multiple values are not there – rst May 12 '16 at 9:43

Let $E$ be a point inside the triangle such that $EB=EC=BC$.

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

• How did you arrive at this construction? Practice? – N.S.JOHN May 12 '16 at 10:49
• @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.) – mathlove May 12 '16 at 11:14

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