Given: ZWZX ZYZX; ZW = ZY Prove: ZX is a perpendicular bisector of WY

Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter3: Triangles
Section3.2: Corresponding Parts Of Congruent Triangles
Problem 43E
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We know that angle WZX is congruent to angle YZX and that sides ZW is congruent to side
ZY. The
property allows us to say that line segment ZX is
congruent to itself. Thus by the
congruency theorem, AWZX
E AYZX. We know that Corresponding Parts of Congruent Triangles are
, so LZXY = LZXW. Because they make up the straight line
segment WY, the two angles are a
which means that mzZXY + mzZXW =
°. Because the angles are equal, we can use substitution to
get 2(M ZXW) = 180°. Solving, we find that both angles equal 90°. This implies that line
segment ZX is
to WY by the definition of perpendicular.
Finally, using CPCTC again we know that WX = YX. Therefore, X is the
of WY, and ZX is the
of WY.
Transcribed Image Text:Fill in the blanks We know that angle WZX is congruent to angle YZX and that sides ZW is congruent to side ZY. The property allows us to say that line segment ZX is congruent to itself. Thus by the congruency theorem, AWZX E AYZX. We know that Corresponding Parts of Congruent Triangles are , so LZXY = LZXW. Because they make up the straight line segment WY, the two angles are a which means that mzZXY + mzZXW = °. Because the angles are equal, we can use substitution to get 2(M ZXW) = 180°. Solving, we find that both angles equal 90°. This implies that line segment ZX is to WY by the definition of perpendicular. Finally, using CPCTC again we know that WX = YX. Therefore, X is the of WY, and ZX is the of WY.
- 1 atta
Given: ZWZX ZYZX; ZW a ZY
Prove: ZX is a perpendicular bisector of WY
Transcribed Image Text:- 1 atta Given: ZWZX ZYZX; ZW a ZY Prove: ZX is a perpendicular bisector of WY
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