ABC is an isosceles right angle ,right angle at B prove that AC²=2 AB²
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Step-by-step explanation:
Given data :
Δ ABC is a right Δ and also isosceles triangle.,
To prove:
\mathrm{AB}^{2}=2 \mathrm{AC}^{2}
Step 1:
Proof:
Here,
Hypotenuse = AB
Also, as it is given that, ΔABC is isosceles,
Step 2:
AC = BC [equal sides of isosceles Δ]
Using Pythagoras theorem,
Step 3:
In Δ ABC, we have ;
\begin{array}{l}{\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2}} \\ {\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{AC}^{2}}\end{array} [AC = BC]
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Step-by-step explanation:
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