Compare and contrast the Angle Angle Similanty Postulate, the Side Side Side Similarity Theorem and the Side Angle Side Similanty Theorem
Write about the similantes and the differences that you observe.?

The similarities are

1) Two triangles are similar when they meet either the (Angle Angle) AA, (Side Side Side) SSS or (Side Angle Side) SAS criteria

2) When two triangles meet either of the above similarity criteria they automatically meet the other similarity criteria

3) The ratio of their equivalent sides are equal such that when ΔABC is similar to ΔDCE we have;

AB/DC = AC/DE = BC/CE

The observed differences are

1) Triangles that meet the SAS and SSS Similarity Theorem criteria can be said to be congruent, that is they have both the same side sizes and angle sizes while triangles that meet only the AA Similarity Postulate criteria may or may not be congruent

2) The number of possible triangles formed by the SAS or SSS Similarity Theorem criteria is only one while the number of possible triangles formed by the AA Similarity Postulate criteria is infinite

3) A triangle that meets either the SAS or SSS Similarity Theorem criteria also meets the AA Similarity Postulate criteria

4) A triangle that meets either the AA Similarity Postulate criteria does not necessarily meet the AA Similarity Postulate criteria.

Step-by-step explanation:

The similarity postulates are;

The Angle Angle Similarity Postulate also known as AA

The Side Side Side Similarity Theorem also known as SSS

The Side Angle Side Similarity Theorem also known as SAS

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3