nayelialvarez8643 nayelialvarez8643

23-04-2020
Mathematics

contestada

Suppose ABCD is a rhombus such that the angle bisector of ∠ABD meets

AD at point K. Prove that m∠AKB = 3m∠ABK. Help meee!! What do I put in the box shown below??

Respuesta :

thaovtp1407 thaovtp1407

26-04-2020

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

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