geometryproblem solvingtriangles
In $\triangle ABC$, $AB = AC$ and $\measuredangle B = 40^\circ$ . $D$ is a point on $AB$ produced such that $AD = BC$. Join $DC$. Find $\angle DCB$ .
What I Tried: Here is a picture :-
For this problem, I think that I have to find the use of the information that $AD = BC$ , but I cannot find how. I tried angle-chasing, looked for similar triangles and cyclic quadrilaterals, but could not find any which can be that useful.
Can anyone help?
Let $\omega$, $O$ be the circumcircle and circumcenter of $\triangle ABC$, respectively. Let $P,Q,R,S$ be four points on the shorter arc $AC$ of $\omega$ dividing this arc into five equal parts.
First, we shall prove that $\triangle RSD$ is equilateral. Let $D'$ be a point inside $\omega$ such that $\triangle RSD'$ is equilateral. Also, let $E$ be inside $\omega$ such that $\triangle PQE$ is equilateral. Invoking symmetries we see that $\triangle D'SC \equiv \triangle D'RQ \equiv \triangle EQR \equiv \triangle EPA$. Note that $\angle EQR = \angle QRD'=\angle QRS-60^\circ = 168^\circ - 60^\circ = 108^\circ$. Hence $\angle D'QR = 90^\circ - \frac 12\angle QRD' = 36^\circ$ and $\angle EQD'=108^\circ - 36^\circ = 72^\circ$. But also $\angle D'EQ = 180^\circ - \angle EQR = 180^\circ - 108^\circ = 72^\circ$. Hence $ED'Q$ is isosceles with $QD'=ED'$. Again, using symmetries we see that $AED'C$ is an isosceles trapezoid with $AE=ED'=D'C$. We have $\angle ACD'=\angle SCD' - \angle SCA = 36^\circ - 24^\circ = 12^\circ$. Since $AED'C$ is an isosceles trapezoid, it is cyclic and since $AE=ED'=D'C$, it follows that $\angle D'AC = \frac 12 \angle EAC = \frac 12 \angle ACD'=6^\circ$. Hence $D'$ coincides with $D$.
Now comes my favourite part. Some angle chasing shows that $\angle QCE = 18^\circ = \angle DCB$ and $\angle DQC = 24^\circ = \angle BQE$. Hence $D$ and $E$ are isogonal conjugates in $\triangle BQC$. It follows that $\angle CBD = \angle EBQ$.
Choose $T$ on $\omega$ so that $BT$ is a diameter. Clearly, $\triangle BQE$ is symmetric to $\triangle TRD$ with respect to perpendicular bisector of $QR$. In particular, $\angle RTD = \angle EBQ$.
Let $RT$ intersect $BC$ at $X$. Since $\angle CBD = \angle EBQ = \angle RTD$, quadrilateral $BDXT$ is cyclic. Hence $\angle BDT = \angle BXT$. Then some angle chasing shows that $\angle DOB = 102^\circ = \angle BXT = \angle BDT$. This precisely means that the circumcircle of $DOT$ is tangent to $BD$ at $D$. Tangent-secant theorem yields $BD^2=BO\cdot BT = BO \cdot 2BO = 2BO^2$. Hence $$\frac{BD}{AB} = \frac{BD}{BO} = \sqrt 2,$$ as desired.
$GE=1/2*CE (opposite 30), ACE isosceles (angles 70,70),draw perpendicular to CE, there are 2 congruent right triangles , angle 20, common hypotenuse. So, GE=EI.
Best Answer
Draw $\angle EAD = \angle ADE = 60^0$. So, $\angle CAE = 40^0$.
$\triangle ACE \cong \triangle BAC$ (by side, side, angle).
As $\triangle ACE$ is isosceles with $AC = CE$, and $\triangle ADE$ is equilateral, with common side $AE$, $CD$ is perpendicular to $AE$.
$\angle BCD = 90^0 - \angle CGF = 90^0 - 80^0 = 10^0 \,$ (as $\angle CGA = 100^0$).