Definitions [3]
- Trapezium: A trapezium is a quadrilateral where only two sides are parallel to each other.
Parallelogram: A parallelogram is a quadrilateral whose opposite sides are parallel.
Rhombus: A rhombus is a quadrilateral with four equal-length sides and opposite sides parallel to each other.
Theorems and Laws [2]
Prove that the bisectors of the interior angles of a rectangle form a square.
Given: A rectangle ABCD in which AR, BR, CP, DP are the bisects of ∠A, ∠B, ∠C, ∠D, respectively forming quadrilaterals PQRS.
To prove: PQRS is a square.
Proof:
In Δ ARB,
∠RAB + ∠RBA + ∠ARB = 180°
45° + 45° + ∠ARB = 180°
90° + ∠ARB = 180°
∠ARB = 180° - 90°
∴ ∠ARB = 90°
Similarly, ∠SRQ = 90°
In Δ ARB,
AR = BR ...(i)
ΔASD ≅ Δ BQC ...[By ASA rule]
AS = BQ ...(ii) [by CPCTC]
(i) - (ii)
AR - AS = BR - BQ
SR = RQ ...(iii)
Also, SP = PQ ...(iv)
PQ = RS ...(v)
Hence, PQRS is a square.
The diagonals of a square are perpendicular bisectors of each other.
Given: ABCD is a square, where AC and BD is a diagonal bisect each other at a Point 'O'.
To Prove: ∠AOD = ∠COD = 90°.
Proof:
ABCD is a square whose diagonals meet at O. ......(Given)
OA = OC. ......(Since the square is a parallelogram)(1)
In ΔAOD and ∆COD,
OD = OD .........(Common side)
OA = OC .........(From 1)
AD = DC ..........(All the sides of square have equal length.)
By SSS congruency condition,
∆AOD ≅ ∆COD
Therefore, m∠ AOD = m∠ COD ......(C.A.C.T.)
Since, m∠ AOD and m∠ COD are a linear pair,
∠AOD = ∠COD = 90°.
Hence Proved.
