Question

Arătați că triunghiul ABC este isoscel, unde:​

Answer 1

$a)~~AB=\sqrt{(x_{A}-x_{B})²+(y_A-y_B)²}\\AB=\sqrt{(2+1)²+(3+1)²}\\AB=\sqrt{9+16}\\AB=5\\\\BC=\sqrt{(x_{B}-x_{C})²+(y_B-y_C)²}\\BC=\sqrt{(-1-6)²+(-1-0)²}\\BC=\sqrt{49+1}\\BC=5\sqrt{2}\\\\AC=\sqrt{(x_A-x_C)²+(y_A-y_C)²}\\AC=\sqrt{(2-6)²+(3-0)²}\\AC=\sqrt{16+9}\\AC=5\\\implies AB=AC=isoscel\\\\\\\\ \\ \\ b)~~AB=\sqrt{(x_{A}-x_{B})²+(y_A-y_B)²}\\AB=\sqrt{(2+\sqrt{3}-\sqrt{3})²+(1+1)²}\\AB=\sqrt{2²+2²}\\AB=2\sqrt{2}\\\\BC=\sqrt{(x_{B}-x_{C})²+(y_B-y_C)²}\\BC=\sqrt{(\sqrt{3}+1)²+(-1+\sqrt{3})²}\\BC=\sqrt{3+2\sqrt{3}+1+1-2\sqrt{3}+3} \\BC=\sqrt{3+1+1+3}\\BC=\sqrt{8}\\BC=2\sqrt{2}\\\\AC=\sqrt{(x_{A}-x_{C})²+(y_A-y_C)²}\\AC=\sqrt{(2+\sqrt{3}+1)²+(1+\sqrt{3})²}\\AC=\sqrt{(3+\sqrt{3})²+1+2\sqrt{3}+3}\\AC=\sqrt{9+6\sqrt{3}+3+1+2\sqrt{3}+3}\\AC=\sqrt{16+8\sqrt{3}}\\AC=\sqrt{(2+2\sqrt{3})²}\\AC=2+2\sqrt{3}\\\implies AB=BC=isoscel$