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ABC is an isosceles triangle right angled at C. Prove that \(AB^2 = 2AC^2\).
Answer :
Given, \(\triangle\) ABC is an isosceles triangle right angled at C.
In \(\triangle\) ACB, Angle C = 90°
AC = BC (By isosceles triangle property)
\(AB^2 = AC^2 + BC^2\) [By Pythagoras theorem]
= \(AC^2 + AC^2\) [Since, AC = BC]
\(AB^2 = 2AC^2\)