In the diagram, ABCD is a square with four congruent sides and four right angles. R, S, T, and U are the midpoints of the sides of ABCD. Also, RT ISU and SV= VU. R Prove: A BSR is congruent to A DUT

Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
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Chapter3: Triangles
Section3.4: Basic Constructions Justified
Problem 38E
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2DU = DA
2BR = BA
2DT = DC
Substitution Property
2BS 2DU
2BR = 2DT
Division Property
7)
8)
Definition Property of Congruence Segments
Segment BS is congruent to segment DU
Segment BR is congruent to Segment DT
Right Angles Congruence Theorem
9)
Angle B is congruent Angle D
10)
Transcribed Image Text:2DU = DA 2BR = BA 2DT = DC Substitution Property 2BS 2DU 2BR = 2DT Division Property 7) 8) Definition Property of Congruence Segments Segment BS is congruent to segment DU Segment BR is congruent to Segment DT Right Angles Congruence Theorem 9) Angle B is congruent Angle D 10)
C
In the diagram, ABCD is a square with four congruent
sides and four right angles. R, S, T, and U are the
midpoints of the sides of ABCD. Also, RT I SU and
V
SV= VU.
A
Prove: A BSR is congruent to A DUT
田
Statement
Reason
1)
The sides of ABCD are congruent, R, S, T, and
U are midpoints of the sides ABCD; angle B
and angle D are right angles.
2)
Definition of Segment Congruence
3)
BS = SC
DU = UA
BR = RA
DT = TC
4)
Segment Addition Postulate
5)
2BS = BC
2DU = DA
2BR = BA
2DT = DC
Substitution Property
6)
2BS = 2DU
2BR = 2DT
Transcribed Image Text:C In the diagram, ABCD is a square with four congruent sides and four right angles. R, S, T, and U are the midpoints of the sides of ABCD. Also, RT I SU and V SV= VU. A Prove: A BSR is congruent to A DUT 田 Statement Reason 1) The sides of ABCD are congruent, R, S, T, and U are midpoints of the sides ABCD; angle B and angle D are right angles. 2) Definition of Segment Congruence 3) BS = SC DU = UA BR = RA DT = TC 4) Segment Addition Postulate 5) 2BS = BC 2DU = DA 2BR = BA 2DT = DC Substitution Property 6) 2BS = 2DU 2BR = 2DT
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