<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>
Answer: In particular, let’s focus our attention on the behavior of each graph at and around . 2 and x= -1 for x < 2. There are open circles at both endpoints (2, 1) and (-2, 1). The third is h (x) = 1 / (x-2)^2, in which the function curves asymptotically towards y=0 and x=2 in quadrants one and two."
Step-by-step explanation: I think this is the problem ur on
2/9 • 3/4 • 2/5 = 1/15
1/15 of of the students are girl sopranos in chorus.
Step-by-step explanation:
-3+m/9=10
m/9=10+3
m/9=13
m=13×9
m=117
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