The figure below shows a rectangle abcd having diagonals ac and db: a rectangle abcd is shown with diagonals ac and bd. jimmy wrote the following proof to show that the diagonals of rectangle abcd are congruent: jimmy's proof: statement 1: in triangle adc and bcd, ad = bc (opposite sides of a rectangle are congruent) statement 2: angle adc = angle bcd (angles of a rectangle are 90°) statement 3: statement 4: triangle adc and bcd are congruent (by sas postulate) statement 5: ac = bd (by cpctc) which statement below completes jimmy's proof?

Statement 3: DC=CD (Common)

Step-by-step explanation:

According from the  statement given, consider ΔADC and ΔBCD, we get

AD=BC(opposite sides of a rectangle are congruent)

∠ADC=∠BCD(angles of a rectangle are 90°)

DC=CD(Common)

Thus, by SAS rule, ΔADC≅ΔBCD and by CPCTC AC=BD.

Thus, Statement 3 that is using the common sides of the triangle jimmy can prove the proof.

Answer

Statement 3: The side BC = BC (common side for ΔADC and ΔBCD)

Solution

Rectangle ABCD having diagonals AC and DB:  

proof

Statement 1: In triangle ADC and BCD, AD = BC (opposite sides of a rectangle are congruent)

Statement 2: Angle ADC = Angle BCD (angles of a rectangle are 90°)  

Statement 3: The side BC = BC (common side for ΔADC and ΔBCD)

Statement 4: Triangle ADC and BCD are congruent (by SAS postulate) Statement 5: AC = BD (by CPCTC)

The statement 3 completes Jimmy's proof.

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step-by-step explanation: