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

_____

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.

The major difference is that a histogram is only used to plot the frequency of score occurrences in a continuous data set that has been divided into classes, called bins.  bar charts, on the other hand, can be used for a great deal of other types ofvariables  including  ordinal  andnominal  data sets.