Question

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

Answer 1

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