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Macedonian Mathematical Competitions

Geometry Difficulty 5.4 AIME, harder Prove it North Macedonia

Let $p$ be a line through the vertex $D$ of a given parallelogram $ABCD$ , such that the vertices $A$ , $B$ and $C$ are on the same side of $p$ . Let $A'$ , $B'$ and $C'$ be the bases of the altitudes from $A$ , $B$ and $C$ to $p$ correspondingly. Prove that $\overline{BB'} = \overline{AA'} + \overline{CC'}$ .

Solution

We draw a line parallel to $p$ through $A$ and let this line intersect $BB'$ in $Q$ . The quadrangle $AQB'A'$ has three right angles, hence it is a rectangle, from where we obtain that $\overline{AA'} = \overline{QB'}$ ....(1).

Because $ABCD$ is a parallelogram, we have that $\overline{AB} = \overline{DC}$ and $\overline{AB} \parallel \overline{CD}$ . The angles $\angle QAB$ and $\angle C'DC$ are equal. Because $BB'$ and $CC'$ are perpendicular to the line $p$ , we get that they are parallel. Hence the angles $\angle QBA$ and $\angle C'CD$ are equal. Because $\overline{AB} = \overline{DC}$ , $\angle QAB = \angle C'DC$ and $\angle QBA = \angle C'CD$ , we conclude that $\triangle ABQ \cong \triangle DCC'$ . Hence $\overline{CC'} = \overline{BQ}$ ....(2).

From (1) and (2) we get $\overline{BB'} = \overline{QB'} + \overline{BQ} = \overline{AA'} + \overline{CC'}$ .

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