Question

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

`\angle ABK\cong \angle DBK` (Definition of angle bisector)

`m\angle ABK=m\angle DBK` (Definition of concurrency)

Let as assume the measure of angle ABK is x.

`m\angle ABK=x`

`m\angle ABD=m\angle ABK+m\angle DBK`

`m\angle ABD=x+x`

`m\angle ABD=2x`

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.

`m\angle ADB=m\angle ABD`

`m\angle ADB=2x`

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

`m\angle AKB=m\angle ADB+m\angle DBK`

`m\angle AKB=2x+x`

`m\angle AKB=3x`

`m\angle AKB=3(m\angle ABK)`

Hence proved.

Answer 2

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.