Suppose ABCD is a rhombus such that the angle bisector of ∠ABD meets AD at point K. Prove that m∠AKB = 3m∠ABK. ?

1 Answer
Mar 12, 2018

drawn

Given:

#ABCD# is a rhombus such that the angle bisector of #angleABD# meets AD at point K.

RTP:

#mangleAKB=3mangle ABK#

Proof:

#BK# is bisector of #angleABD#

So #nangle ABK=mangle KBD=x ^@ (say)#

Hence #mangle ABD=2x^@#

Now #AB = AD#, two sides of rhombus #ABCD#

Hence #mangle KDB=mangleABD=2x^@#

#angle AKB# being exterior angle of #DeltaKBD#, we can write

#mangleAKB=mangleKDB+mangle KBD=2x^@+x^@=3x^@ #

Hence #mangleAKB=3mangle ABK#, proved