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Let $ABC$ be an isosceles triangle with $AB=AC$. Equilateral triangles are constructed on the sides of the triangle. Let $BCP$ and $ACQ$ be external equilateral triangles and $R$ be the point where $AP$ and $BQ$ intersect. Let $ABU$ and $BCV$ internal triangles constructed inwards $ \triangle ABC$ and $W$ the point of intersection of $AV$ and $CU$. Show $RD= DW$ where $D$ is the foot of the altitude from $A$ to $BC$.

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There are two possible configurations, that is one of those.

Clearly, $A,V,R,D$ and $W$ are collinear, since the perpendicular bisector of $BC$ goes through all of them.

$ \angle VBA= \angle CBU = \angle ACV = 60 - \angle B= 60 - \angle C$

$ \angle A= 180 - 2\angle B= 180 - 2( 60 - \angle ACV)= 60 + 2\angle ACV$

$ \angle ABQ = \angle AQB = \frac {120 - \angle A}{2}= 30 - \angle ACV$

Then:

$ \angle VBA + \angle ABQ + \angle RBD= \angle ACV + \angle ABQ + \angle RBD = 60$

$ \angle ACV + (30 - \angle ACV) + \angle RBD = 60$

$ \angle RBD = 30$

So to complete the problem we just have to show that $ \angle RBD= DCW=30$ or to show that $QB$ is parallel to $CW$. I did several things but failed to show that.

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One of my ideas was to extend $BA, CB$ and $BW$ until the intersect $CV, BV$ and $CP$ at $G,E$ and $H$ at respectively. Then: $ \angle VEC = \angle VGB = \angle BHC$ and $ \angle BHC + \angle BGC = \angle BHC + BEC = 180 $ and therefore $B,C, H , G $ and $ E$ lie on a circle. That was the only thing I could figure out. Can you help me finish please. Thanks in advance.

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$$BU=AB=AC=CQ,$$ Also, we have: $$\Delta ABC\cong \Delta AUQ,$$ which gives $$BC=UQ.$$ Thus, $BQCU$ is a trapezoid, $$BQ||CU$$ and from here $D$ is a midpoint of $RW.$

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