320 is on the Fourth quadrant:)
<span>Consider a angle â BAC and the point D on its defector
Assume that DB is perpendicular to AB and DC is perpendicular to AC.
Lets prove DB and DC are congruent (that is point D is equidistant from sides of an angle â BAC
Proof
Consider triangles ΔADB and ΔADC
Both are right angle, â ABD= â ACD=90 degree
They have congruent acute angle â BAD and â CAD( since AD is angle bisector)
They share hypotenuse AD
therefore these right angle are congruent by two angle and sides and, therefore, their sides DB and DC are congruent too, as luing across congruent angles</span>
There could be 2 possible shapes that could’ve been drawn.
A rectangle, or a square.
All of their angles are 90° angles, which are right angles.
Hope this helped!
-8
f(x) is y, in which x is the independent variable
when x = 5, y = -8
