Question

given: ( be = df )( ce = ef )prove: ( bc = de )statement1 ( be = bc + ce )2 ( df = de + ef )3 ( be = df )4 ( bc + ce = de + ef )5 ( ce = ef )reasonreason 2reason 2reason 2reason 2reason 2

Explanation:

Step1: Segment addition postulate

By the segment addition postulate, \(BE = BC + CE\) (because \(B - C - E\) are collinear) and \(DF=DE + EF\) (because \(D - E - F\) are collinear).

Step2: Substitute given equalities

Given \(BE = DF\) (statement 3), we substitute \(BE\) with \(BC + CE\) and \(DF\) with \(DE + EF\) (from step 1) to get \(BC + CE=DE + EF\).

Step3: Substitute \(CE = EF\)

Given \(CE = EF\) (statement 5), substitute \(EF\) with \(CE\) in the equation \(BC + CE=DE + EF\). So \(BC + CE=DE + CE\).

Step4: Subtract \(CE\) from both sides

Subtract \(CE\) from both sides of the equation \(BC + CE=DE + CE\). Using the subtraction property of equality (\(a + c=b + c\Rightarrow a = b\)), we get \(BC=DE\).

Answer:

1. Reason 1: Segment Addition Postulate
2. Reason 2: Segment Addition Postulate
3. Reason 3: Given
4. Reason 4: Substitution Property of Equality
5. Reason 5: Given
6. Reason 6: Substitution Property of Equality (substitute \(EF\) with \(CE\) in \(BC + CE=DE + EF\))
7. Reason 7: Subtraction Property of Equality (\(BC + CE-CE=DE + CE - CE\))