Question

In the adjoining figure, AD = DE. Prove that AB + BC > CЕ.

Answer 1

Given:

• AD = DE and points A, B, C, D, E are placed as in the figure D lies on AC, DE is the segment from D down to E.

To Prove:

• AB + BC > CE

Proof [Step-wise]:

1. In triangle ABC, by the triangle inequality, the sum of two sides is greater than the third side.

Hence, AB + BC > AC.

2. In triangle CDE, again by the triangle inequality CE ≤ CD + DE equality would occur only if C, D, E are collinear.

3. From the given AD = DE,

Substitute DE by AD in step 2 to get CE ≤ CD + AD.

But CD + AD = AC.

Since D lies on AC. 

Therefore, CE ≤ AC.

4. Combine the inequalities from steps 1 and 3:

AB + BC > AC and CE ≤ AC. 

Thus, AB + BC > CE.