<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>
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there u go if that helps u can copy and paste it

Actually Welcome to the Concept of the Similarities.
Since here, first figure is a Triangle, hence similarity of a triangle is given by ratio of it sides and make them equate,
hence for,
1.) 16/12 = x/9 ===> 4/3 = x/9 ===> x = 9*4/3 ==> x = 36/3
hence the value of x is 12 , ===> x = 12 .
2.) here,in the case of a Reactangle the relation is such that, length = 2* breadth,
hence we apply the relation,
x = 2*4.5 ===> x = 9 units.
Answer:

Step-by-step explanation:
Subtract 7 From both sides

7-7 = 0, 16 - 7=9
That leaves 
Hope this helps
~R3V0