Question

Given WXYZ is a rhombus.

Answer 1

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,

$WX=WZ$        (sides of rhombus are equal)

$XY=ZY$        (sides of rhombus are equal)

$WY=WY$       (Common side)

By SSS postulate,

$\triangle WXY\cong \triangle WZY$

$\angle WYX\cong \angle WYZ$       (CPCTC)

$\angle XWY\cong \angle ZWY$       (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,

$WX=YX$        (sides of rhombus are equal)

$XZ=XZ$         (Common side)

$WZ=YZ$        (sides of rhombus are equal)

By SSS postulate,

$\triangle WXZ\cong \triangle YXZ$

$\angle WZX\cong \angle YZX$       (CPCTC)

$\angle WXZ\cong \angle YXZ$       (CPCTC)

Hence prove, that ZX bisects Angle WZY and Angle YXW.