Hl suppose abcd is a rhombus such that the angle bisector of ∠abd meets ad at point k. prove that m∠akb = 3m∠abk.

1. Let m∠ABK=x°. Since line BK bisects ∠ABD, then

m∠ABK=m∠KBD=x°.

Also m∠ABD=m∠ABK+m∠KBD=2x°.

2. The diagonal BD of rhombus ABCD bisects ∠ABC, then

m∠ABD=m∠DBC=2x°.

This gives you that

m∠ABC=4x°.

3. Angles A and B are supplementary, so

m∠A+m∠B=180°,

m∠A=180°-4x°.

4. Consider triangle ABK. The sum of the measures of interior angles in triangle is always 180°, thus

m∠A+m∠ABK+m∠AKB=180°,

m∠AKB=180°-x°-(180°-4x°),

m∠AKB=3x°=3m∠ABK.

Given information: ABCD is a rhombus, angle bisector of ∠ABD meets AD at point K.

        (Definition of angle bisector)

         (Definition of concurrency)

Let as assume the measure of angle ABK is x.

All sides of a rhombus are same.

Since AB=AD, therefore ABD is an isosceles triangle and two angles of an isosceles triangle are congruent.

According to exterior angle theorem, sum of two interior angles of a triangle is equal to third exterior angle.

Using exterior angle theorem, we get

Hence proved.

maybe - definition of supplementary angles. im not sure. sorry

step-by-step explanation: