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

Mar 12, 2018

Given:

$A B C D$ is a rhombus such that the angle bisector of $\angle A B D$ meets AD at point K.

RTP:

$m \angle A K B = 3 m \angle A B K$

Proof:

$B K$ is bisector of $\angle A B D$

So $n \angle A B K = m \angle K B D = {x}^{\circ} \left(s a y\right)$

Hence $m \angle A B D = 2 {x}^{\circ}$

Now $A B = A D$, two sides of rhombus $A B C D$

Hence $m \angle K D B = m \angle A B D = 2 {x}^{\circ}$

$\angle A K B$ being exterior angle of $\Delta K B D$, we can write

$m \angle A K B = m \angle K D B + m \angle K B D = 2 {x}^{\circ} + {x}^{\circ} = 3 {x}^{\circ}$

Hence $m \angle A K B = 3 m \angle A B K$, proved