Question

ABC is an isosceles triangle with AC = BC. If AB² = 2AC², Prove that ABC is a right triangle.

Answer 1

Explanation:

I hope it will be help you

Answer 2

The converse of the Pythagorean theorem.

Triangle must satisfy $\sf{\overline{AB}^2=\overline{BC}^2+\overline{CA}^2}$ to become a right triangle.

One of the given condition is $\sf{\overline{BC}=\overline{CA}}$

We get $\sf{\overline{BC}^2=\overline{CA}^2}$

Another given condition is $\sf{\overline{AB}^2=2\overline{CA}^2}$

$\sf{\implies \sf{\overline{AB}^2=\overline{CA}^2+\overline{CA}^2}}$

$\sf{\implies \overline{AB}^2=\overline{BC}^2+\overline{CA}^2}$ (Right-angled at $\sf{\angle{C}}$)

So we have a right triangle according to the converse of the Pythagorean theorem.

Hence proven.

Interesting facts

$\sf{\overline{AB}}$ is √2 times the other sides.

$\sf{\implies\overline{AB}=\sqrt{2} \;\overline{BC}}$

$\sf{\implies\overline{AB}=\sqrt{2} \;\overline{CA}}$

The triangle area is $\dfrac{1}{4}$ times the square of the hypotenuse.

Pythagorean theorem was not proven by himself. It was Euclid who proven it for the first time.

Fermat's last theorem is similar to $\sf{a^2+b^2=c^2}$, but any integer satisfy over the power of 2. One of the interesting facts is Fermat didn't prove it. 358 years since Andrew Wiles came up with the perfect proof.