Respuesta :

thaovtp1407

Answer:

Step-by-step explanation:

As we know that:

  • BK is bisector of ∠ABD

=> ∠ABK=∠KBD=x

=> ∠ABD=2x

  • Now AB=AD, two sides of rhombus ABCD

=> ∠KDB=∠ABD=2x

  • ∠AKB being exterior angle of ΔKBD, we have:

∠AKB=∠KDB+∠KBD=2x+x=3x

Please have a look at the attached photo

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