Which algebraic expression os a polynomial with a degree of 2
1 answer:
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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,
(sides of rhombus are equal)
(sides of rhombus are equal)
(Common side)
By SSS postulate,
(CPCTC)
(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,
(sides of rhombus are equal)
(Common side)
(sides of rhombus are equal)
By SSS postulate,
(CPCTC)
(CPCTC)
Hence prove, that ZX bisects Angle WZY and Angle YXW.
