Isosceles Triangle and Altitude:

An isosceles triangle is a triangle that has at least two sides of equal length. In an isosceles triangle, the altitude from the vertex (the point where the two equal sides meet) to the base (the side opposite the vertex) bisects the base. This means that the altitude divides the base into two equal segments.

Proof:

To prove that the altitude from the vertex bisects the base in an isosceles triangle, we can use the properties of congruent triangles and the perpendicular bisector.

Step 1: Draw an isosceles triangle ABC with AB = AC.

Step 2: Draw the altitude from vertex A to the base BC and label the point of intersection as D.

Step 3: Now, we need to prove that BD = DC.

Step 4: Since triangle ABC is isosceles with AB = AC, we know that angle ABC = angle ACB.

Step 5: The altitude AD is perpendicular to BC, which means that angle BDA = angle CDA = 90 degrees.

Step 6: Using the Angle-Angle (AA) postulate, we can conclude that triangle ABD is congruent to triangle ACD (angle BDA = angle CDA and angle ABD = angle ACD).

Step 7: By the definition of congruence, we know that corresponding sides of congruent triangles are equal. Therefore, we have BD = CD.

Step 8: Hence, the altitude from the vertex A bisects the base BC into two equal segments, BD and CD.

Conclusion:

In an isosceles triangle, the altitude from the vertex bisects the base. This can be proved using the properties of congruent triangles and the perpendicular bisector. The proof shows that the two segments created by the altitude are equal, thus confirming that the altitude bisects the base.