Answer:
We need to prove that the angle bisector of the angle opposite the base of an isosceles triangle is also the median and altitude to the base.
Let's consider an isosceles triangle ABC where AB = AC. We draw the altitude from A to BC and call the point where it intersects BC as D.
Now, we need to prove that AD is the angle bisector, median, and altitude to the base BC.
To prove AD is the angle bisector:
We need to prove that the angle ADB and ADC are equal. We know that angle ABD and angle ACD are right angles because BD and CD are altitudes. We also know that AB = AC because the triangle is isosceles. Therefore, the triangles ABD and ACD are congruent by the hypotenuse-leg (HL) criterion.
Thus, angle ADB = angle ADC, which means that AD is the angle bisector of angle BAC.
To prove AD is the median:
We need to prove that BD = CD. Since AB = AC and AD is perpendicular to BC, triangles ABD and ACD are congruent by the hypotenuse-leg (HL) criterion. Therefore, BD = CD, which means that AD is also the median to the base.
To prove AD is the altitude:
We need to prove that angle BAD and angle CAD are right angles. This is true because AD is perpendicular to BC, and BD and CD are also perpendicular to BC. Therefore, AD is also the altitude to the base BC.
Hence, we have proved that the angle bisector of the angle opposite the base of an isosceles triangle is also the median and altitude to the base.