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Given: KLMN is a parallelogram,

KA

− angle bisector of ∠K

LA

− angle bisector of ∠L

Prove: m∠KAL = 90°

answer in format LKM+?= (something degrees) by reason ?

Answer :

Answer:

∠ KAL = 90° (Proved)

Step-by-step explanation:

See the attached diagram.

As KLMN is a parallelogram, so ∠K + ∠ L = 180°

[tex]\frac{1}{2}\angle K + \frac{1}{2}\angle L = 90^{\circ}[/tex] ............ (1)

Now, given that, [tex]\angle AKL = \frac{1}{2} \angle K[/tex] and [tex]\angle ALK = \frac{1}{2} \angle L[/tex]

So, ∠ AKL + ∠ ALK = 90° {From equation (1)}

Now, from Δ KAL, ∠ AKL + ∠ ALK + ∠ KAL = 180°

⇒ 90° + ∠ KAL = 180°

⇒ ∠ KAL = 90° (Proved)

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