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

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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