### 1977 IMO SL #8

Let $S$ be a convex quadrilateral $ABCD$ and $O$ a point inside it. The feet of the perpendiculars from $O$ to $AB, BC, CD, DA$ are $A_1, B_1, C_1, D_1$ respectively. The feet of the perpendiculars from $O$ to the sides of $S_i$, the quadrilateral $A_iB_iC_iD_i$, are $A_{i+1}B_{i+1}C_{i+1}D_{i+1}$, where $i = 1, 2, 3.$ Prove that $S_4$ is similar to S.

Note that $OB_1BA_1$ is cyclic, so $\angle OBC = \angle OA_1B_1$. Furthermore, we have $\angle{OA_4B_4}=\angle{OB_3C_3}=\angle{OC_2D_2}=\angle{OD_1A_1}=\angle{OAB}$ and similarly we have $\angle{OBA}=\angle{OB_4A_4}$. Hence, combining these two, we get $\triangle{OA_4B_4}\sim \triangle {OAB}$. Doing the same for the other triangles, we get $$\triangle OC_4D_4 \sim \triangle OCD, \triangle OD_4A_4 \sim \triangle ODA, \triangle OB_4C_4 \sim \triangle OBC, \triangle OA_4B_4 \triangle OAB.$$ Using all of these, we get $ABCD \sim A_4B_4C_4D_4$, so we are done. $\square$

### Comments

1. Anonymous2/27/2022

wonderful solution!

2. Anonymous2/28/2022

geometry is my favorite :)