Problem — Proportionality of medians, bisectors and altitudes in similar triangles

Text

Prove that in two similar triangles the corresponding medians, bisectors and altitudes are in the same ratio as the homologous sides.

Solution

Let $ABC$ and $A'B'C'$ be two similar triangles with similarity ratio $k=\frac{\overline{A'B'}}{\overline{AB}}=\frac{\overline{B'C'}}{\overline{BC}}=\frac{\overline{C'A'}}{\overline{CA}},$ and with congruent corresponding angles: $\hat A\cong\hat A'$, $\hat B\cong\hat B'$, $\hat C\cong\hat C'$.

Altitudes. Let $AH$ and $A'H'$ be the altitudes to the sides $BC$ and $B'C'$. The right triangles $ABH$ and $A'B'H'$ have $\hat B\cong\hat B'$ and the right angles at $H$ and $H'$: hence they are similar (first criterion). Therefore $\frac{\overline{A'H'}}{\overline{AH}}=\frac{\overline{A'B'}}{\overline{AB}}=k.$

Bisectors. Let $AD$ and $A'D'$ be the bisectors of the angles $\hat A$ and $\hat A'$. Since $\hat A\cong\hat A'$, their halves are congruent: $\widehat{BAD}\cong\widehat{B'A'D'}$. Then triangles $ABD$ and $A'B'D'$ have $\hat B\cong\hat B'$ and $\widehat{BAD}\cong\widehat{B'A'D'}$, hence they are similar and $\frac{\overline{A'D'}}{\overline{AD}}=\frac{\overline{A'B'}}{\overline{AB}}=k.$

Medians. Let $AM$ and $A'M'$ be the medians to $BC$ and $B'C'$, with $M$ and $M'$ the midpoints. Then $\frac{\overline{BM}}{\overline{B'M'}}=\frac{\tfrac12\overline{BC}}{\tfrac12\overline{B'C'}}=\frac{\overline{BC}}{\overline{B'C'}}=\frac{\overline{AB}}{\overline{A'B'}},$ and moreover $\hat B\cong\hat B'$: triangles $ABM$ and $A'B'M'$ have the included angle congruent and the sides about it proportional, hence (second criterion) they are similar and $\frac{\overline{A'M'}}{\overline{AM}}=\frac{\overline{A'B'}}{\overline{AB}}=k.$

In all three cases the ratio between corresponding segments equals the ratio $k$ of the homologous sides. $\blacksquare$