Given : NL bisects KNM and KLM Prove: ΔNKL ≅ ΔNML

ASAP NEED HELP Assignment DUE IN 10 MINS

ΔNKL ≅ ΔNML by ASA rule of congruency

Step-by-step explanation:

The given parameters are;

Given the quadrilateral, KLMN, with diagonal = LN

The diagonal LN = Angle bisector of angles ∠KNM and ∠KLM

Therefore, we have;

∠KNM = ∠KNL + ∠MNL and ∠KNL ≅ ∠MNL by definition of angles in angles formed by angle bisector NL

Given;

∠KNL ≅ ∠MNL by angles formed by an angle bisector angle

≅ by reflexive property

∴ ΔNKL ≅ ΔNML by Angle-Side-Angle (ASA) rule of congruency

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

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step-by-step explanation: