Please help. The question is in the image.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

answer: the two figures are congruent.

step-by-step explanation: they're congruent because figure efgh is reflected across the x and y axis but it was also rotated.

hope it !

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

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

step-by-step explanation: