Name: Problem Based on Midpoint Theorem
Uploaded: 2019-07-19
Duration: 5 min 16 s
Description: The line segment in a triangle joining the midpoint of two sides of the triangle is parallel to its third side and is also half of the length of the third side.

Question

Show that the diagonals of a square are equal and right bisectors of each other.

Accepted Answer

Given that $A B C D$ is a square.

To prove: $A C = B D$ and, $A C$ and $B D$ bisect each other at right angles.

Proof: In a $Δ A B C$ and $Δ B A D$

$A B = A B$ ( common side)

$B C = A D$ ( opposite sides of a square)

$∠ A B C = ∠ B A D = 90 °$

$Δ A B C ≅ Δ B A D$ ( By $S A S$ congruence)

$A C = B D$ (by Corresponding parts of the congruent triangle).

Now in $Δ O A D$ and $Δ O C B$

$A D = C B$ (opposite sides of a square)

$∠ O A D = ∠ O C B$ (alternate interior angles are equal)

$∠ O D A = ∠ O B C$ (alternate interior angles are equal)

$Δ O A D ≅ Δ O C B$ (by $A S A$ property)

$O A = O C$

Similarly, $O B = O D$ ...............(i)

From (i) and (ii) $A C$ and $B D$ bisect each other.

Now in a $Δ O B A$ and $Δ O D A$

$O B = O D$ (from (i))

$B A = D A$ (sides of a square are equal)

$O A = O A$ (common side)

$∆ O B A ≅ ∆ O D A$ (by $S S S$ )

$∠ A O B = ∠ A O D$ (Corresponding parts of the congruent triangle)

$∠ A O B + ∠ A O D = 180 °$ (linear pair)

$2 ∠ A O B = 180 °$

$∠ A O B = 90 ° = ∠ A O D$

Therefore, $A C$ and $B D$ bisect each other at right angles.

Show that the diagonals of a square are equal and right bisectors of each other.

Important Questions on Quadrilaterals

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