Yes
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No. A median is a line from a vertex to the midpoint of the opposite side. It divides the triangle into congruent parts only if
the triangle is equilateral
or
if the triangle is isosceles and it is the median from the unequal vertex.
In all other cases the two parts will not be congruent.
Let the triangle be ABC and suppose the median AD is also an altitude.AD is a median, therefore BD = CDAD is an altitude, therefore angle ADB = angle ADC = 90 degreesThen, in triangles ABD and ACD,AD is common,angle ADB = angle ADCand BD = CDTherefore the two triangles are congruent (SAS).And therefore AB = AC, that is, the triangle is isosceles.
they are the same because the triangles side is equal
median
2/3 of the median is between the centroid and the vertex, 1/3 between the centroid and the side.
Sure. That's true of a median in every isosceles triangle, and every median in an equilateral triangle. In fact it is true for any median of any triangle. The two parts may not be the same shapes but they will have the same area. That is why the point where the three medians meet (centroid) is the centre of mass of a triangular lamina of uniform thickness.
It is the line joining the midpoints of two sides of a polygon - usually a triangle. This line will be parallel to the third side. The three median-median lines will divide any triangle into 4 congruent triangles that are similar to the original.It is the line joining the midpoints of two sides of a polygon - usually a triangle. This line will be parallel to the third side. The three median-median lines will divide any triangle into 4 congruent triangles that are similar to the original.It is the line joining the midpoints of two sides of a polygon - usually a triangle. This line will be parallel to the third side. The three median-median lines will divide any triangle into 4 congruent triangles that are similar to the original.It is the line joining the midpoints of two sides of a polygon - usually a triangle. This line will be parallel to the third side. The three median-median lines will divide any triangle into 4 congruent triangles that are similar to the original.
In the diagram, ABC is an isoscels triangle with the congruent sides and , and is the median drawn to the base . We know that ∠A ≅ ∠C, because the base angles of an isosceles triangle are congruent; we also know that ≅ , by definition of an isosceles triangle. A median of a triangle is a line segment drawn from a vertex to the midpoint of the opposite side. That means ≅ . This proves that ΔABD ≅ ΔCBD. Since corresponding parts of congruent triangles are congruent, that means ∠ABD≅ ∠CBD. Since the median is the common side of these adjacent angles, in fact bisects the vertex angle of the isosceles triangle.
Let the triangle be ABC and suppose the median AD is also an altitude.AD is a median, therefore BD = CDAD is an altitude, therefore angle ADB = angle ADC = 90 degreesThen, in triangles ABD and ACD,AD is common,angle ADB = angle ADCand BD = CDTherefore the two triangles are congruent (SAS).And therefore AB = AC, that is, the triangle is isosceles.
Isosceles.
A median divides any triangle in half.
they are the same because the triangles side is equal
median
Medians bisect the sides of ALL triangles. That is what a median is, by definition!
For the equilateral triangle in Euclidean space(i.e, the triangles you see in general) median is the same as its altitude. So, both are of equal length.
2/3 of the median is between the centroid and the vertex, 1/3 between the centroid and the side.
The altitude is the segment from an angle of a triangle to the side opposite of the angle which is intersected perpendicularly by the altitude., the angle bisector cuts an angle into two congruent angles, and a median forms two congruent line segments.
Sure. That's true of a median in every isosceles triangle, and every median in an equilateral triangle. In fact it is true for any median of any triangle. The two parts may not be the same shapes but they will have the same area. That is why the point where the three medians meet (centroid) is the centre of mass of a triangular lamina of uniform thickness.