Euler olympiad 2009 P3 In the triangle $\triangle ABC$, the sides $AB$ and $BC$ are equal. The point $D$ inside the triangle is such that the angle $\angle ADC$ is twice as large as the angle $\angle ABC$. Prove that the double distance from the point $B$ to the line bisecting the angles externally to the angle $\angle ADC$ is $AD+DC$.
I am not able to understand what it's asking, and what we have to prove so, please help. 
 A: I invite you, as an exercise, to fill in the missing details. 
Consider the Figure below.



*

*Let $\gamma$ be the circumcircle of $\triangle ABC$ and $O$ its center. $D$ must lie on the circumcircle $\gamma'$ of $\triangle AOC$. Why?

*Let $C'$ be the intersection of $AD$ with $\gamma$. Demonstrate that $\triangle DCC'$ is isosceles. As a consequence $D$ lies on the axis of the chord $CC'$. Conclude that $OD$ is the external bisector of $\angle ADC$.

*Draw from $B$ the line perpendicular to $OD$, that intersects $\gamma$ in $B'$ and $OD$ in $H$. Can you demonstrate that $BB'CC'$ is an isosceles trapezoid? Thus conclude that $BC'\cong B'C$.

*Use the above information to demonstrate that $\triangle BB'C \cong \triangle BAC'$, and, in particular, that $BB' \cong AC'$.

*Recalling that $\triangle DCC'$ is isosceles and that $BB' \cong 2BH$, show that $$2BH \cong AD + DC,$$ as desired.


Hints for point 1.

 This is not strictly necessary to solve the problem, but it is useful to draw a picture. Given a segment $AB$ and two points $C$ and $C'$ on the same half-plane formed by $AB$, then $\angle ACB \cong \angle AC'B$ if and only if $ABCC'$ is cyclic.

Hints for point 2.

 $\angle AC'C \cong \angle ABC$ (they both subtend chord $AB$ on $\gamma$), and thus $\angle ADC \cong 2 \angle AC'C$. From the External Angle Theorem we have $\angle ADC \cong \angle DC'C + \angle DCC'$, and thus $\triangle DCC'$ is isoseles. Since $DC'\cong DC$ and also $OC' \cong OC$, both $O$ and $D$ lie on the axis of $CC'$, which is also the bisector of $\angle CDC'$.

Hints for point 3.

 $BB'\parallel CC'$, because they are both perpendicular to $OD$. Opposite angles of a cyclic quadrilateral are supplementary, which implies that $BB'CC'$ is isosceles. (Widely speaking: any cyclic trapezoid is isosceles.)

Hints for point 4.

 Use the congruence of angles subtended by congruent chords in $\gamma$ plus the AAS criterion.

A: Most of the work are suggested by @dfnu . I have added the bold-faced part to clarify why X, the center of the circle ABC, lies on HD.
AD extended and BH extended cut circle $\omega$ at C’ and B’ respectively.
Let the external and the internal angle bisector of $\angle ADC$ cut $\omega$ at M and D’ respectively. Then, $\angle D’DM = 90^0$ and therefore $BB’ // DD’$.

$\angle 1 = \angle 2 = \angle 3$ implies $C’C // DD’$. This further means $\triangle DNC \cong \triangle DNC’$; where $N = HDM \cap CC’$. Then, MNDH is the perpendicular bisector of CC’.
Let MNDH cut the circle passing through ADC at X. The just mentioned two lines will meet at most two points, namely D and X. Either X or D is the center of $\omega$. We will skip the latter case (because it becomes an over-simplified special case of the question.)
$XH \bot BB’$ implies $2BH = BB’$.
By equal chord implies equal angle at circumference, $\alpha = \alpha '$, and $\beta = \beta '$. Together with $BC’ = B’C$, we have $\triangle BB’C \cong \triangle AC’B$. Required result follows. 
