138. На рис. 43 АК = AM, CK = CM. Доказать, что KO = OM.

Решение:

Рассмотрим треугольники \( \triangle AKO \) и \( \triangle AMO \). У нас есть:

• \( AK = AM \) (по условию).
• \( AO = AO \) (общая сторона).
• \( \angle KAO = \angle MAO \) (так как \( AK = AM \) и \( CK = CM \) imply that \( AO \) is the angle bisector of \( \angle KCM \) and also \( AO \) is the angle bisector of \( \angle KAM \) is not explicitly stated but can be inferred if \( \triangle AKC \) and \( \triangle AMC \) are isosceles and \( AO \) bisects the vertex angle, which is the case here because \( AK=AM \) and \( CK=CM \) means \( \triangle AKC \) and \( \triangle AMC \) are congruent by SSS if we assume C is on the line passing through A and O, which is not given. Let's assume \( \angle AKO = \angle AMO \) is not given). Instead, we can use the property of isosceles triangles.

  Consider \( \triangle AKC \) and \( \triangle AMC \). Since \( AK = AM \) and \( CK = CM \), and \( AC \) is common, \( \triangle AKC \) is congruent to \( \triangle AMC \) by SSS congruence if C is a vertex. However, we are given \( CK = CM \). This suggests \( \triangle AKC \) and \( \triangle AMC \) are not necessarily congruent unless \( AC \) is a common side, which it is.

  Let's restart with a clearer approach using the given conditions.

  We are given \( AK = AM \) and \( CK = CM \). This means that point \( A \) and point \( C \) are on the perpendicular bisector of the segment \( KM \) if \( AK = CK \) and \( AM = CM \). This is not given.

  However, if \( AK = AM \) and \( CK = CM \), then \( \triangle AKM \) is an isosceles triangle with \( AK = AM \). Also, \( \triangle CKM \) is an isosceles triangle with \( CK = CM \).

  Consider \( \triangle AKO \) and \( \triangle AMO \).

  We have \( AK = AM \) (given).

  We have \( AO = AO \) (common side).

  Let's consider \( \triangle CKO \) and \( \triangle CMO \).

  We have \( CK = CM \) (given).

  We have \( CO = CO \) (common side).

  This does not directly help us prove \( KO = OM \).

  Let's rethink the problem using the properties of isosceles triangles.

  Since \( AK = AM \), \( \triangle AKM \) is isosceles. The line segment \( AO \) might be related to the angle bisector of \( \angle KAM \).

  Since \( CK = CM \), \( \triangle CKM \) is isosceles. The line segment \( CO \) might be related to the angle bisector of \( \angle KCM \).

  If \( A, O, C \) are collinear, and \( AO \) is the angle bisector of \( \angle KAM \) and \( CO \) is the angle bisector of \( \angle KCM \), then \( AO \) and \( CO \) are medians and altitudes of \( \triangle AKM \) and \( \triangle CKM \) respectively. If \( A, O, C \) are collinear, then \( AC \) is a line passing through the vertices \( A \) and \( C \) of the isosceles triangles.

  Consider the case where \( A, O, C \) are collinear. Then \( AC \) is a line of symmetry for \( \triangle AKM \) and \( \triangle CKM \) IF \( AO \) and \( CO \) are angle bisectors from the vertex angle to the base. For \( \triangle AKM \) with \( AK=AM \), \( AO \) being the angle bisector of \( \angle KAM \) implies \( AO \) is also the median to \( KM \), so \( KO=OM \). Similarly for \( \triangle CKM \) with \( CK=CM \), if \( CO \) is the angle bisector of \( \angle KCM \), then \( CO \) is also the median to \( KM \), so \( KO=OM \).

  The problem statement implies that \( A, O, C \) are points on a line and \( O \) is on \( KM \) or somewhere in relation to \( K \) and \( M \). From the figure, it appears that \( O \) is the intersection of \( AC \) and \( KM \).

  Given \( AK = AM \), \( \triangle AKM \) is isosceles. Let \( AO \) be the angle bisector of \( \angle KAM \). Then \( KO = OM \). This is a property of isosceles triangles: the angle bisector from the vertex angle is also the median and the altitude to the base.

  Given \( CK = CM \), \( \triangle CKM \) is isosceles. Let \( CO \) be the angle bisector of \( \angle KCM \). Then \( KO = OM \). This is also a property of isosceles triangles.

  If we assume that \( AO \) is the angle bisector of \( \angle KAM \) and \( CO \) is the angle bisector of \( \angle KCM \), then \( O \) lies on the segment \( KM \) and \( KO = OM \). The conditions \( AK = AM \) and \( CK = CM \) imply that \( A \) and \( C \) are on the perpendicular bisector of \( KM \) only if \( AK = CK \) and \( AM = CM \) which is not given. However, it means that \( A \) is equidistant from \( K \) and \( M \), and \( C \) is equidistant from \( K \) and \( M \).

  Consider \( \triangle AKC \) and \( \triangle AMC \). We have \( AK = AM \) and \( CK = CM \). If \( AC \) is a line segment, and \( O \) is a point on \( AC \) such that \( KO \) and \( OM \) are line segments, then the diagram suggests that \( O \) is the intersection of \( AC \) and \( KM \).

  Let's prove that \( AO \) is the angle bisector of \( \angle KAM \) and \( CO \) is the angle bisector of \( \angle KCM \).

  Consider \( \triangle AKM \). Since \( AK = AM \), it is an isosceles triangle. Let \( AO \) be the line segment from \( A \) to \( O \) on \( KM \). If \( AO \) is the angle bisector of \( \angle KAM \), then \( KO = OM \).

  Consider \( \triangle CKM \). Since \( CK = CM \), it is an isosceles triangle. Let \( CO \) be the line segment from \( C \) to \( O \) on \( KM \). If \( CO \) is the angle bisector of \( \angle KCM \), then \( KO = OM \).

  The fact that \( AK = AM \) means \( A \) lies on the perpendicular bisector of \( KM \) only if \( AK = AM \) and \( BK = BM \) for some other points \( B \) on the line. In this case, \( AK = AM \) means that \( A \) is on the angle bisector of \( \angle KXM \) where \( X \) is some point.

  Let's use the property that if two sides of a triangle are equal, then the line from the vertex to the midpoint of the base is also the angle bisector and altitude. If \( O \) is the midpoint of \( KM \), then \( KO = OM \). We need to prove that \( AO \) and \( CO \) are angle bisectors.

  Given \( AK = AM \) and \( CK = CM \). This implies that point \( K \) and point \( M \) are symmetric with respect to the line \( AC \). Thus, the line \( AC \) is the perpendicular bisector of \( KM \). Therefore, the intersection of \( AC \) with \( KM \), which is point \( O \), must be the midpoint of \( KM \). Hence, \( KO = OM \).

  Proof:

  1. Given \( AK = AM \). This means that \( A \) is equidistant from \( K \) and \( M \). In terms of locus, \( A \) lies on the perpendicular bisector of the segment \( KM \).

  2. Given \( CK = CM \). This means that \( C \) is equidistant from \( K \) and \( M \). In terms of locus, \( C \) lies on the perpendicular bisector of the segment \( KM \).

  3. Since both \( A \) and \( C \) lie on the perpendicular bisector of \( KM \), the line passing through \( A \) and \( C \) is the perpendicular bisector of \( KM \).

  4. Let \( O \) be the intersection of the line \( AC \) and the segment \( KM \). Since the line \( AC \) is the perpendicular bisector of \( KM \), the point \( O \) must be the midpoint of \( KM \).

  5. Therefore, \( KO = OM \).

  This completes the proof.

  Ответ: Доказано.