Answer:
Diagonal of rhombus bisect the angles of it.
Step-by-step explanation:
Given: WXYZ is a rhombus.
To prove: WY bisects Angle ZWX and Angle XYZ. ZX bisects Angle WZY and Angle YXW.
In rhombus WXYZ, WY is a diagonal.
In triangle WXY and triangle WZY,
[tex]WX=WZ[/tex] (sides of rhombus are equal)
[tex]XY=ZY[/tex] (sides of rhombus are equal)
[tex]WY=WY[/tex] (Common side)
By SSS postulate,
[tex]\triangle WXY\cong \triangle WZY[/tex]
[tex]\angle WYX\cong \angle WYZ[/tex] (CPCTC)
[tex]\angle XWY\cong \angle ZWY[/tex] (CPCTC)
Hence prove, that WY bisects Angle ZWX and Angle XYZ.
Similarly,
In rhombus WXYZ, ZX is a diagonal.
In triangle WXZ and triangle YXZ,
[tex]WX=YX[/tex] (sides of rhombus are equal)
[tex]XZ=XZ[/tex] (Common side)
[tex]WZ=YZ[/tex] (sides of rhombus are equal)
By SSS postulate,
[tex]\triangle WXZ\cong \triangle YXZ[/tex]
[tex]\angle WZX\cong \angle YZX[/tex] (CPCTC)
[tex]\angle WXZ\cong \angle YXZ[/tex] (CPCTC)
Hence prove, that ZX bisects Angle WZY and Angle YXW.
