# External and internal equilateral triangles constructed on the sides of an isosceles triangles, show…

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 $$BC$$.

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.

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.

$$BU=AB=AC=CQ,$$ 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 $$RW.$$