Proof:

Let's consider two isosceles triangles with a common base.

Given:
- Two isosceles triangles with a common base.
- In an isosceles triangle, the base angles are equal.

To Prove:
- The line segment joining the vertices of the triangles bisects their common base at right angles.

Proof:

Step 1: Draw the isosceles triangles and their common base.

Draw two isosceles triangles ABC and ABD with a common base AB.

A
/ \
/ \
/ \
B-------C
\ /
\ /
\ /
D

Step 2: Mark the equal angles.

Since ABC and ABD are isosceles triangles, the base angles will be equal. Mark the angles ∠BAC and ∠BAD as equal.

A
/ \
/ ∠ \
/ \
B-------C
\ /
\ ∠ /
\ /
D

Step 3: Connect the vertices of the triangles.

Draw a line segment CD joining the vertices C and D.

A
/ \
/ ∠ \
/ \
B-------C
\ /
\ ∠ /
\ /
D

Step 4: Prove that CD bisects AB at right angles.

To prove that CD bisects AB at right angles, we need to show that ∠CDA = ∠CDB = 90°.

Proof of ∠CDA = 90°:
- In triangle ABC, ∠BAC = ∠BAD (given).
- The sum of angles in a triangle is 180°.
- Therefore, ∠BAC + ∠ABC + ∠CBA = 180°.
- Since ABC is an isosceles triangle, ∠ABC = ∠CBA.
- Substituting ∠BAC = ∠BAD and ∠ABC = ∠CBA, we get 2∠BAD + ∠BAD + ∠BAD = 180°.
- Simplifying the equation, we have 4∠BAD = 180°.
- Dividing both sides by 4, we get ∠BAD = 45°.
- Since ∠BAD = ∠CDA (alternate angles), ∠CDA = 45°.

Proof of ∠CDB = 90°:
- In triangle ABD, ∠ABD = ∠ADB (given).
- The sum of angles in a triangle is 180°.
- Therefore, ∠ABD + ∠BAD + ∠ADB = 180°.
- Since ABD is an isosceles triangle, ∠ABD = ∠ADB.
- Substituting ∠ABD = ∠ADB and ∠BAD = 45°, we get ∠ADB + 45° + ∠ADB = 180°.
- Simplifying the equation, we have 2∠ADB + 45° = 180°.
- Subtracting 45° from both sides, we get 2