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12 December, 14:14

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.

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  1. 12 December, 14:59
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    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.
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