Answer:
Step-by-step explanation:
Hope this helps u !!
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.
Answer:
I think the answer is B
Step-by-step explanation:
sorry if Im wrong.
PLZ MARK BRAINLIEST
Answer:
the vertices are all labeled
a b c d and e
Answer:
28 should do it
Step-by-step explanation:
35+35+82= 152
180-152=28