Question

please answer as fast as you can ​

Answer 1

Answer:

if bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isoscelesif bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isoscelesif bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isoscelesif bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isoscelesif bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isoscelesif bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isoscelesif bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isosceles bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isosceles is isoscelesif bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isosceles

if bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isoscelesif bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isoscelesif bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isoscelesif bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isoscelesif bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isoscelesif bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isoscelesif bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isosceles bisector AE of exterior vertical angle DAE of triangle abc be parallel to base BC prove that tringe is isosceles is isoscele

${xyb \beta \beta \gamma \cos( log_{ log_{ \beta }(?) }(?) ) }^{?}$