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Mathematics

1 answer:

asambeis [7]2 years ago

3 0

Answer:

<DEF = 107

Step-by-step explanation:

The way to best solve this problem is by starting with angle F. Remember that all angles in a straight line must add up to 180. So by subtracting 135 from 180 we get 45 for angle F.

180 - 135 = 45

After that, we can figure out the measure of <DEF. All the angles of a triangle must add up to 180. So:

180 - 28 - 45 = 107

Making the answer 107 for the measure of <DEF

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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.

Neporo4naja [7]

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.