Question

In the adjacent figure, BE = DE. Prove that AC + BC > CD.

Answer 1

Given:

• A, B, C, D, E are as in the figure, with AB vertical, BE the segment on AB, DE horizontal and BE = DE.

To Prove:

• AC + BC > CD.

Proof (Step-wise):

1. Consider triangle CBD.

By the triangle inequality, the sum of two sides is greater than the third, we have CD < CB + BD.

2. Consider triangle ABC. 

By the triangle inequality again,

AC + CB > AB

3. Combine (1) and (2).

To show AC + CB > CD it suffices to show that the right-hand side of (2) (which is > AB) is strictly larger than BD less whatever shortfall remains, but we will proceed by bounding BD in terms of AB and BE = DE to get a strict inequality.

4. In right triangle BED (DE ⟂ BE because DE is horizontal and BE is vertical) with BE = DE.

BD is the hypotenuse,

So, `BD = BE xx sqrt(2)`.

Since `sqrt(2) < 2` we have BD < 2 × BE. 

Also, BE is a proper part of AB.

So, BE < AB.

5. From step 4, we have BD < 2 × BE < 2 × AB.

Therefore, CB + BD < CB + 2 × AB.

6. From step 2, we already have AC + CB > AB.

Adding AB to both sides of this inequality gives (AC + CB) + AB > 2 × AB.

Hence, AC + CB > 2 × AB – AB = AB.

Combining this with BD < 2 × AB (from step 4) and CB ≥ 0, we get AC + CB > AB and CB + BD < CB + 2 × AB.

Taking the comparison between the two sums and using (1) yields AC + CB > CD.

AC + BC > CD, as required.