In the adjoining figure, XY = XZ . YQ and ZP are the bisectors of [tex] \angle[/tex] XYZ and [tex] \angle[/tex] XZY respectively. Prove that YQ = ZP.
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![In The Adjoining Figure, XY = XZ . YQ And ZP Are The Bisectors Of [tex] \angle[/tex] XYZ And [tex] \angle[/tex]](https://brainimage.s3.us-east-005.backblazeb2.com/contents/attachments/30qMjiXACddTBGM9obS1hrNyOJqk1wk5.jpeg)
Answers
Answer:
See Below.
Step-by-step explanation:
Statements: Reasons:
[tex]1)\, XY=XZ[/tex] Given
[tex]2) \text{ $ m\angle Y= m\angle Z$}[/tex] Isosceles Triangle Theorem
[tex]\displaystyle 3) \text{ $m\angle Y=m\angle XYQ + \angle QYZ$}[/tex] Angle Addition
[tex]\displaystyle 4)\text{ $YQ$ bisects $\angle XYZ$}[/tex] Given
[tex]5) \text{ $m\angle XYQ=m\angle QYZ$}[/tex] Definition of Bisector
[tex]\displaystyle 6)\text{ $m\angle Y=2m\angle QYZ$}[/tex] Substitution
[tex]7)\text{ $m\angle Z=m\angle XZP+m\angle PZY$}[/tex] Angle Addition
[tex]8)\text{ $ZP$ bisects $\angle XZY$}[/tex] Given
[tex]\displaystyle 9) \text{ $m\angle XZP=m\angle PZY$ }[/tex] Definition of Bisector
[tex]\displaystyle 10) \text{ $ m\angle Z = 2m\angle PZY $}[/tex] Substitution
[tex]11)\text{ } 2m\angle QYZ=2m\angle PZY[/tex] Substitution
[tex]12)\text{ }m\angle QYZ=m\angle PZY[/tex] Division Property of Equality
[tex]13)\text{ } YZ=YZ[/tex] Reflexive Property
[tex]14)\text{ } \Delta YZP\cong\Delta ZYQ[/tex] Angle-Side-Angle Congruence*
[tex]15)\text{ } YQ=ZP[/tex] CPCTC
*For clarification:
∠Y = ∠Z
YZ = YZ (or ZY)
∠PZY = ∠QYZ
So, Angle-Side-Angle Congruence:
ΔYZP is congruent to ΔZYQ
Answer:
this is your answer. look at this once.
![In The Adjoining Figure, XY = XZ . YQ And ZP Are The Bisectors Of [tex] \angle[/tex] XYZ And [tex] \angle[/tex]](https://brainimage.s3.us-east-005.backblazeb2.com/answers/attachments/GFtlllDMZoCZmDqMZNG6jOzswMb9OVi1.jpeg)
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