Question

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??

Answer 1

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