Suppose abcd is a rhombus and that the bisector of ∠abd meets
ad
at point k. prove that m∠akb = 3m∠abk.
m∠akb = m∠kbd + m∠
by reason
find the angle that missing angle so that angle kbd and that angle will equal angle akb.
explain

missing angle: ∠DBC

Step-by-step explanation:

Proof:

m∠ABK ≅ m∠KBD — given that BK bisects ∠ABDm∠ABD = m∠ABK + m∠KBD = 2·m∠ABKm∠ABD ≅ m∠DBC — properties of a rhombus: a diagonal bisects the anglesm∠DBC = 2·m∠ABK — transitive property (both equal to m∠ABD)m∠KBC = m∠KBD + m∠DBC — adjacent anglesm∠KBC = m∠ABK + 2·m∠ABK = 3·m∠ABK — substitute for m∠KBD and m∠DBCm∠AKB = m∠KBC — alternate interior angles of parallel lines AD, BCm∠AKB = 3·m∠ABK — substitute for m∠KBC

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Proof is always in the eye of the beholder, and the details depend on the supporting theorems and postulates you're allowed to invoke. The basic idea is that you have cut a vertex angle in half twice, and you're trying to show that the smallest part to the rest of it has the ratio 1 : 3.

Just tell koda to adjust the number of shapes himself.