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)
