Question

Prove that in any parallelogram the sum of the squares on the diagonals is twice the sum of the squares on two adjacent sides.

Answer 1

Answer on Question #43545, Math, Other

Task: Prove that in any parallelogram the sum of the squares on the diagonals is twice the sum of the squares on two adjacent sides.

Answer:

We have parallelogram ABCD with diagonals AC and BD. Using the Law of Cosines:

$$\Delta ABC: AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos \angle B;$$ $$\text{In } \Delta ABD: BD^2 = AB^2 + AD^2 - 2 \cdot AB \cdot AD \cdot \cos \angle A;$$

But $AD = BC; \angle A = 180{}^\circ - \angle B$.

$$BD^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(180{}^\circ - \angle B) \Rightarrow$$

So we have $BD^2 = AB^2 + BC^2 + 2 \cdot AB \cdot BC \cdot \cos \angle B; \Rightarrow$

$$AC^2 + BD^2 = 2(AB^2 + BC^2)$$

www.AssignmentExpert.com