Answer:
This is proved by ASA congruent rule.
Step-by-step explanation:
Given KLMN is a parallelogram, and that the bisectors of ∠K and ∠L meet at A. we have to prove that A is equidistant from LM and KN i.e we have to prove that AP=AQ
we know that the diagonals of parallelogram bisect each other therefore the the bisectors of ∠K and ∠L must be the diagonals.
In ΔAPN and ΔAQL
∠PNA=∠ALQ (∵alternate angles)
AN=AL (∵diagonals of parallelogram bisect each other)
∠PAN=∠LAQ (∵vertically opposite angles)
∴ By ASA rule ΔAPN ≅ ΔAQL
Hence, by CPCT i.e Corresponding parts of congruent triangles PA=AQ
Hence, A is equidistant from LM and KN.
The correct transformation is a rotation of 180° around the origin followed by a translation of 3 units up and 1 unit to the left.
<h3>
Which transformation is used to get A'B'C'?</h3>
To analyze this we can only follow one of the vertices of the triangle.
Let's follow A.
A starts at (3, 4). If we apply a rotation of 180° about the origin, we end up in the third quadrant in the coordinates:
(-3, -4)
Now if you look at A', you can see that the coordinates are:
A' = (-4, -1)
To go from (-3, -4) to (-4, -1), we move one unit to the left and 3 units up.
Then the complete transformation is:
A rotation of 180° around the origin, followed by a translation of 3 units up and 1 unit to the left.
If you want to learn more about transformations:
brainly.com/question/4289712
#SPJ1
Step-by-step explanation:
-2x=30-120
-2x = -90
=x = -90/-2
=x = +45
