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In the given figure, ABC is an isosceles right triangle, right-angled at C. Prove that $A{{B}^{2}}=2A{{C}^{2}}$.
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Last updated date: 23rd Jul 2024
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Hint: Two sides of an isosceles triangle is equal. Pythagoras theorem can be given as ${{\left( \text{hypotaneous} \right)}^{\text{2}}}\text{=}{{\left( \text{base} \right)}^{\text{2}}}\text{+}{{\left( \text{perpendicular} \right)}^{\text{2}}}$. Using these two things the desired result can be obtained.

Complete step-by-step answer:
We have an isosceles triangle ABC with$\angle C={{90}^{o}}$ and hence sides AC and BC are equal to each other, as in isosceles triangle two sides are equal. Hence, diagram of $\Delta ABC$can be represented as
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Where AC = BC and $\angle C={{90}^{o}}$
Now, we need to prove the relation given as
$A{{B}^{2}}=2A{{C}^{2}}$ ………………….. (i)
So, let us calculate the value of $A{{B}^{2}}$in terms of $A{{C}^{2}}$to prove the above equation (i).
As we know any right angle triangle will follow the Pythagoras property which can be given as
${{\left( \text{hypotaneous} \right)}^{\text{2}}}\text{=}{{\left( \text{base} \right)}^{\text{2}}}\text{+}{{\left( \text{perpendicular} \right)}^{\text{2}}}$………… (iii)
Hence, we can write the equation (iii) in terms of sides of $\Delta ABC$ as
${{\left( AB \right)}^{2}}={{\left( BC \right)}^{2}}+{{\left( AC \right)}^{2}}$…………… (iv)
Now, as the given triangle is isosceles, therefore AC=BC from the figure; and hence, we can replace BC by AC in the equation (iv), so, we get
 & {{\left( AB \right)}^{2}}={{\left( AC \right)}^{2}}+{{\left( AC \right)}^{2}} \\
 & \Rightarrow {{\left( AB \right)}^{2}}=2{{\left( AC \right)}^{2}} \\
Hence, equation (i) or the given relation is proved.

One can go wrong with the Pythagoras theorem. One may put the value of base or perpendicular in place of hypotenuse or vice-versa may also happen. Hence, be clear with the terms of the Pythagoras theorem.
One can get angles A and B as ${{45}^{o}}$. As both are equal and summation of all the angles of the triangle is ${{180}^{0}}$. Now, take $\sin {{45}^{o}}$in the triangle with respect to angle B.
Hence $\sin B=\sin {{45}^{o}}=\dfrac{AC}{AB}$
Now, put $\sin {{45}^{o}}=\dfrac{1}{\sqrt{2}}$ and square both sides.
We will get the same result as given in the question.