Please help me with this problem

ΔAOK ≅ ΔBOL by SAS

Step-by-step explanation:

Hello there!

We are given:

Line segment AB intersects line segment KL at point O

O is the midpoint of AB

O is the midpoint of L

We want to prove that ΔAOK is congruent to ΔBOL

From the given picture, line segments AK and LB don't exist, but we can draw them in

Once we've drawn them in, we can use the given information to help us

Since O is the midpoint of AB, AO = OB by the definition of a midpoint

O is also the midpoint of LK, and because of that, LO=OK, also because of the definition of a midpoint

Since O is the intersection of AB and KL, that means that O would help form the vertical angles <LOB and <KOA (which are congruent because of vertical angles theorem)

Now, we have 2 sides and an angle which are congruent.

In the answer, we already have ΔAOK filled in

Remember that when naming congruent triangles, the congruent pieces need to match

A corresponds with B

O corresponds with O

K corresponds with L

So, that means ΔAOK ≅ ΔBOL

The reason is because of SAS (side-angle-side). The reason why it's SAS is because we have 2 pairs of congruent sides and one angle which is congruent

Please see the picture below if you'd like a visual aid

Hope this helps!

side xy = 10

sin(x) = 4/5

cos(x) = 3/5

tan(x) = 4/3

step-by-step explanation:

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions