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.

Answer 1

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.