# Showing that a quadrilateral is a square

euclidean-geometrygeometry

I am trying to prove the following:

"Consider a square ABCD. Draw an external line r through D and call H and K the projections of A and B on r respectively, and with R the projection of A on BK. Show that the quadrilateral AHKR is square."

Now, it is easy to see that $$\hat{R}$$, $$\hat{K}$$ and $$\hat{H}$$ are all right angles since R, K and H are projections.

Now it is left to show that the four sides are congruent but since I haven't managed to do so without using vectors/coordinates I would appreciate an hint about how to do so without using these methods but only using classic euclidean geometry, thanks.

$$AH \perp AR$$

$$AB=AD$$ (Sides of Square)

$$\angle ADH =\angle DAR$$(alt. int $$\angle 's, DH \parallel AR$$)

$$\angle ABR + \angle BAR = \angle DAR + \angle BAR=90^o$$

Hence,

$$\angle ABR = \angle DAR$$ and $$\angle ABR = \angle DAR$$ and

$$\angle ADH =\angle ABR$$

and

$$\angle ADH + \angle DAH = \angle ABR + \angle BAR=90^o$$

Hence, $$\angle DAH =\angle BAR$$

$$\triangle ADH \cong \triangle ABR$$

Therefore $$AH=AR$$

Two consecutive sides both perpendicular and equal.

$$AHKR$$ is a square