Read the proof.
Given: AB I DE
Prove: AACB - ADCE
We are given AB DE. Because the lines are parallel and
segment CB crosses both lines, we can consider segment
CB a transversal of the parallel lines. Angles CED and CBA
are corresponding angles of transversal CB and are
therefore congruent, so _CED CBA. We can state C
C using the reflexive property. Therefore, AACB -
ADCE by the
E
AA similarity theorem.
SSS similarity theorem.
AAS similarity theorem.
ASA similarity theorem.

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it

so, converting firstly the mixed fractions to improper fractions and then "adding".

c

step-by-step explanation:

see below

step-by-step explanation:

x+2> 6

subtract 2 from each side

x+2-2 > 6-2

x > 4

there is an open circle at 4 since is it is greater than and the line goes to the right