<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>
You take
7.25 (10)+5.5p=105.5
72.5+5.5p=105.5
To make the equation easier multiply the whole equation by ten like this
(72.5+5.5p=105.5)10 that equals
725+55p=1055
Then subtract 725 to both sides
725+55p=1055
-725 -725
____ _____
55p=330
Then divide by 55 on both sides and that equals 6 so 6 people bought tickets
1/8
3 divide by 3=1
24 divide by 3= 8
which gives you 1/8
Measure of angle 2 would be 130 degrees because since it forms a linear pair the two angles added together have to equal 180 degrees so
180 - 50 = 130
