Prove the diagonals of a square are perpendicular
1 answer:
Answer:
Given :- ABCD is a square.
To proof :- AC = BD and AC ⊥ BD
Proof :- In △ ADB and △ BCA
AD = BC [ Sides of a square are equal ]
∠BAD = ∠ABC [ 90° each ]
AB = BA [ Common side ]
△ADB ≅ △BCA [ SAS congruency rule ]
⇒ AC = BD [ Corresponding parts of congruent triangles are equal ]
In △AOB and △AOD
OB = OD [ Square is also a parallelogram therefore, diagonal of parallelogram bisect each other ]
AB = AD [ Sides of a square are equal ]
AO = AO [ Common side ]
△AOB ≅ △ AOD [ SSS congruency rule ]
⇒ ∠AOB = ∠AOD [ Corresponding parts of congruent triangles are equal]
∠AOB + ∠AOD = 180° [ Linear pair ]
∠ AOB = ∠AOD = 90°
⇒ AO ⊥ BD
⇒ AC ⊥ BD
Hence proved, AC = BD and AC ⊥BD
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Answer:
Center: (-1,8)
Radius: 1
The graph is attached.
Step-by-step explanation:
The equation of the circle has the form:
Where (h,k) is the point of the center of the circle and r is the radius of the circle.
The equation given in the problem is
Therefore:
h=-1
k=8
Then, the center is (-1,8) and radius is 1.
You can graph the circle with its center at the (-1,8) and a radius of 1 as you can see in the figure attached.
