Question

In the given diagram, AE² + BD² is equal to:

Options

• AB² − DE²

• DE² − BD²

• AB² + DE²

• DE × AB

Answer 1

AB² + DE²

Explanation:

In right-angled triangle ΔACE, applying Pythagoras’ theorem gives AE² = AC² + CE².

Similarly, in right-angled triangle ΔBCD, we have BD² = BC² + CD².

Adding these two equations gives AE² + BD² = (AC² + BC²) + (CE² + CD²).

Since AC² + BC² = AB² in ABC and CE² + CD² = DE² in ΔCDE, substituting these gives:

AE² + BD² = AB² + DE².