Name: Angle Sum Property of a Quadrilateral
Uploaded: 2019-07-19
Duration: 7 min 55 s
Description: We know that quadrilaterals have four sides and four angles. To know more about what makes the angles of a quadrilateral special, involve yourself with this...

Question

Show that the diagonals of a square are equal and bisect each other at right angles.

Accepted Answer

Let $A B C D$ is a square with its diagonals $A C$ and $B D$ intersecting at $O .$

$(a)$ To prove that the diagonals are equal, i.e. $A C = B D$

In $Δ A B C$ and $Δ B A D,$

$AB = BA$ (common side)

$B C = A D$ (opposite sides of the square $A B C D$ ]

$∠ A B C = ∠ B A D$ (angles of a square are equal to $90^{o}$ ]

$∴ Δ A B C ≅ Δ B A D$ (SAS criteria)

$\Rightarrow$ Their corresponding parts are equal.

$\Rightarrow A C = B D . . . . (1)$

$(b)$ To prove that $' O'$ is the midpoint of $A C$ and $B D .$

$∵ A D ∥ B C$ and $A C$ is a transversal. ( $∵$ opposite sides of a square are parallel)

$∴ ∠ OCB = ∠ OAD$ (interior alternate angles)

Similarly, $∠ OBC = ∠ ODA$ (interior alternate angles)

Also, in $Δ O A D$ and $Δ O C B, A D = C B$ (opposite sides of the square)

Hence, $Δ O A D ≅ Δ O C B$ [ASA criteria]

$∴$ Their corresponding parts are equal.

$\Rightarrow O A = O C$ and $OD = OB$

$\Rightarrow O$ is the midpoint of $A C$ and $B D,$ that is the diagonals $A C$ and $B D$ bisect each other at $O . . . . (2)$

$(c)$ To prove that $A C ⊥ B D$

In $Δ O B A$ and $Δ O D A, O B = O D$ (proved)

$B A = D A$ (opposite sides of the square)

$O A = O A$ (common)

$∴ Δ O B A ≅ Δ O D A$ (SSS criteria)

$\Rightarrow$ Their corresponding parts are equal.

$\Rightarrow ∠ A O B = ∠ A O D$

And $∠ A O B$ and $∠ A O D$ form a linear pair.

$∠ A O B + ∠ A O D = 180^{o}$

$∠ A O B = ∠ A O D = 90^{o}$

$\Rightarrow A C ⊥ B D . . . . (3)$

Therefore, from $(1), (2) and (3)$ , we get $A C$ and $B D$ are equal and bisect each other at right angles.

Show that the diagonals of a square are equal and bisect each other at right angles.

Important Questions on Quadrilaterals

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