Final answer:
In an isosceles triangle ∆ABC with base AC and median BM, we can prove that ∆AMD ≅ ∆CMD using the SSS congruence rule by showing that AM = CM, ∠AMD = ∠CMD, and the side DM is common to both triangles.
Step-by-step explanation:
In an isosceles triangle ΔABC with base AC and median BM, point D is on BM. To prove that ΔAMD ≅ ΔCMD, we can use the SSS congruence rule.
- First, we can establish that AM = CM since M is the mid-point of AC, so the medians are congruent.
- Next, we can prove that ∠AMD = ∠CMD. Since ΔABC is isosceles, we know that ∠B = ∠C. Since M is the mid-point of AC, we know that AM is congruent to CM. Therefore, ∠AMD = ∠CMD because they are corresponding angles of congruent triangles.
- Finally, we can conclude that ΔAMD ≅ ΔCMD by the SSS congruence rule because we have proven that AM = CM, ∠AMD = ∠CMD, and DM is common to both triangles.