Proof:

Given: Two isosceles triangles with a common base.

To prove: The line segment joining their vertices bisects the common base at right angles.

Construction:
1. Let's consider two isosceles triangles, ABC and ADC, with a common base AC.
2. Join the vertices B and D.
3. Draw the perpendicular bisector of AC, which intersects BD at point O.

Proof:

Step 1: Prove that AB = AC and AD = AC
1. In triangle ABC, AB = AC (as ABC is an isosceles triangle).
2. In triangle ADC, AD = AC (as ADC is an isosceles triangle).
3. Therefore, AB = AC = AD.

Step 2: Prove that ΔABC and ΔADC are congruent triangles
1. AB = AC (proved in Step 1).
2. AD = AC (proved in Step 1).
3. AC = AC (common side).
4. Therefore, by Side-Side-Side congruence, ΔABC ≅ ΔADC.

Step 3: Prove that angle BOC and angle DOA are right angles
1. As ΔABC ≅ ΔADC (proved in Step 2), angle BAC = angle DAC.
2. Also, angle BCA = angle DCA (as base angles of an isosceles triangle are equal).
3. Therefore, angle BOC = angle DOA (vertical angles).
4. Hence, angle BOC and angle DOA are congruent.

Step 4: Prove that BO = OD
1. In triangle BOC and triangle DOA, BC = DC (as both are the same base).
2. AB = AD (proved in Step 1).
3. Therefore, by Side-Angle-Side congruence, triangle BOC ≅ triangle DOA.
4. Hence, BO = OD (corresponding parts of congruent triangles are equal).

Step 5: Prove that line segment BD bisects AC at right angles
1. BO = OD (proved in Step 4).
2. The perpendicular bisector of a line segment passes through the midpoint of the segment.
3. Therefore, line segment BD bisects AC at point O.
4. Also, angle BOC and angle DOA are right angles (proved in Step 3).
5. Hence, line segment BD bisects AC at right angles.

Therefore, it is proved that the line segment joining the vertices of two isosceles triangles with a common base bisects the common base at right angles.