maths book class-9 ex 8.1 ka question no. 5
Answers
Ello There!
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Question: Show that if the Diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
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Answer
Given,
In ABCD
AO = CO
OB = OD
∠AOB = ∠AOD = ∠COB = ∠COD = 90°
AC = BD
To Prove
ABCD is a square
Proof
Consider ΔAOB and ΔCOB
AO = CO (given)
OB = OD (given)
∠AOB = ∠COB (Vertically Opposite Angles)
∴ ΔAOB ≅ ΔCOB by SAS congruency.
∴ ∠1 = ∠2 (CPCT)
AB║CD (Alternate angles 1 and 2 are equal)
AB = CD (CPCT) → 1
∴ ABCD is a parallelogram
(One pair of opposite sides are equal and parallel)
Consider ΔAOB and ΔAOD
OA = OA (common)
OB = OD (given)
∠AOB = ∠BOC = 90° (given)
∴ ΔAOB ≅ ΔAOD by SAS congruency
AB = AD (CPCT) → 2
Similarly,
AB = BC → 3
From 1, 2 and 3
AB = BC = CD = DA
Since all sides are equal,
ABCD is a rhombus
Now,
Consider ΔDAB and ΔADC
AD = AD (common)
AB = DC (sides of a rhombus are equal)
AC = BD (given)
∴ ΔDAB ≅ ΔADC by SSS congruency
⇒ ∠ADC = ∠DAB (CPCT)
But,
∠ADC + ∠DAB = 180° (co-interior angles)
∴ ∠ADC = ∠DAB = 90° (∠ADC = ∠DAB)
Since all sides are equal and two angles are 90°,
ABCD is a square.
Proved!
(custom figure, might not look good, but thats what I could make)
