Let $E,F$ be points on $BD,BC$ respectively such that $AE \perp BD$ and $AF \perp BC$. As AD is a tangent and B is an external tangential point, so \[ \angle ABF = \angle ABG + \angle CBG = \angle BAD + \angle BDA = 180^{\circ} - \angle ABD = \angle ABE. \] Therefore $\triangle ABE \equiv \triangle ABF$ and hence $AE=AF$.

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I will solve this nice problem by inversion. Consider an inversion with center $A$ and an arbitrary radius. and suppose points $C,B,D$ changes to $C',B',D'$ respectively. It is clear that the problem changes to this one: there are two parallel lines $L_1, L_2$, a circle is tangent to $L_1$ at $B'$, and intersects$L_2$ at $C',D'$ prove that: circum circles of $AB'D',AB'C'$ have equal diameters. To prove this easy fact, u should only know that a trapezoid can inscribed in a circle if and only if it is isosceles. Ok…………………………………………………………………………………………………………………. Draw the circum circle of $AC'B'$, and let it intersect $L_1$ at $E'$. Then$AC'B'E'$ should be an isosceles trapezoid.and because of $C'B'D'$ is an isosceles triangle, $AE$ is parallel to $BD$. And it shows us that $AEBD$ is parallelogram. Then triangles$AEB,ABD$ are equal, and we can conclude That their circum circles have equal diameters too.

This problem is easy and nice :∠ABC=∠CMB+∠BCM=∠NBM+∠BDC=∠ABP+∠BDA;Because l and AC are tangent lines to Γ1;so PA=PB;PAB=<PBA;SO :∠ABC=∠PAB+∠BDA=∠HBA:So, BA is the angle bisector of <HBG; SO:AH=GA;---Q.E.D

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Let $ BD$ intersect the first circle again at $ E$. $ \angle ABC = \angle AED + \angle ADE = 180^o - \angle EAD = \angle EBA$. Thus $ AB$ is the bisector of $ EBC$ and the result follows.

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Define $E = BB \cap AC$ and let $BD$ meet the circle containing $A$ again at $F$. Since $EA = EB$, we have $$\angle ABC = \angle ABE + \angle EBC = \angle BAE + \angle BDC = \angle BAD + \angle BDA = 180^{\circ} - \angle ABD = \angle ABF$$so $A$ lies on the external bisector of $\angle CBD$, which finishes. $\blacksquare$