Ask Question
21 April, 21:46

Suppose KLMN is a parallelogram, and that the bisectors of ∠K and ∠L meet at A. Prove that A is equidistant from LM and KN.

+4
Answers (1)
  1. 22 April, 01:29
    0
    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.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Suppose KLMN is a parallelogram, and that the bisectors of ∠K and ∠L meet at A. Prove that A is equidistant from LM and KN. ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers