In the figure, point b is the midpoint of ac. use the figure to awnser the questions.

a.

i agree with what jeremy stated that it is proven with sas congruence postulate.

let's start with the sides from the sas postulate.

it is given that point b is the midpoint of ac.

that means ab = bc. that is one side congruence, so we need one more.

point b bisects the line ac, which creates a perpendicular bisector bd.

bd is shared on both triangles as a side. we could say bd = bd

now there are two side congruence that would make sas postulate.

all we need to know is the angle from the sas postulate.

going back to the perpendicular bisector from bd, it creates two right angles (90°) on both sides. this is from the perpendicular bisector postulate.

so ∠abd = ∠cbd.

now the sides and angles combined makes up the sas postulate.

b.

if ad = cd, then the triangle would be proved by sss congruence postulate since all the sides are considered congruent.

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

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