, 2003

Solution:

a. Let $B$ and $C$ be the vertices from which the two equal altitudes start, relative respectively to sides $A C$ and $A B$. Computing the area $S$ of the triangle with respect to $A B$ and to $A C$, and calling $h$ the common length of the two altitudes, we have
$$S = \frac{A B \cdot h}{2} = \frac{A C \cdot h}{2}$$
from which $A B = A C$.

b. Let $B$ and $C$ be the vertices from which the two equal medians start, relative respectively to sides $A C$ and $A B$, which meet the midpoints respectively at $N$ and $M$. We also call $O$ the point where the medians meet (the centroid). As is well known, $O$ divides each median into two parts, of which the one containing the vertex is twice as long as the other: but since the two medians have the same length, the two segments $C O$ and $B O$ (which are $2/3$ of the corresponding median) also have the same length, and therefore the triangle $B O C$ is isosceles and the triangles $B O M$ and $C O N$ are congruent (they have two corresponding sides and the included angle of the same measure: $C O$ and $B O$ as stated, likewise $N O$ and $M O$, and the two angles $C \widehat{O} N$ and $B \widehat{O} M$ because they are vertical angles).

But then the angles $B \widehat{C} O$ and $C \widehat{B} O$ have the same measure (they are base angles of an isosceles triangle), as do the angles $O \widehat{C} N$ and $O \widehat{B} M$ (they are corresponding angles of congruent triangles). Hence $B \widehat{C} A = B \widehat{C} O + O \widehat{C} A$ and $C \widehat{B} A = C \widehat{B} O + O \widehat{B} A$ are congruent and $A B C$ is isosceles.

Another proof: b) is a special case of c).

c. By Thales's theorem, the line of segment $B C$ is parallel to that of segment $M N$. But then the angles $B \widehat{C} M$ and $C \widehat{M} N$ (and also $N \widehat{B} C$ and $B \widehat{N} M$) are congruent because they are formed by parallel lines cut by a transversal. If $O$ is the point where segments $B N$ and $C M$ meet, then $C \widehat{O} B$ and $N \widehat{O} M$ are also congruent because they are vertical angles, so the triangles $B O C$ and $M O N$ are similar. But then
$$B O : O N = C O : O M$$
from which, adding in the proportion,

$$B N : O N = C M : O M$$
But, since $B N = C M$, also $O N = O M$, that is, the triangle $M O N$ (and therefore also the triangle $B O C$) is isosceles. But then the triangles $M B C$ and $N C B$ are congruent (they have $C M = B N$, $B C$ in common, and the included angles $N \widehat{B} C$ and $M \widehat{C} B$ congruent), and therefore the angles $B \widehat{C} A$ and $A \widehat{B} C$ are congruent and the triangle $A B C$ is isosceles.