How i solve the question of class 9th ncert math of Quadrilaterals chapter of exercise 8.1 of question no 4 think it is very important question of class 9th give me the tricky way for solving this question Show that the diagonals of a square are equal and bisect each other at right angles.
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Show that the diagonals of a square are equal and bisect each other at right angles. Q.4
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Let ABCD be a square and its diagonals AC and BD intersect each other at O.
To show that,
AC = BD
AO = OC
and ∠AOB = 90°
Proof,
In ΔABC and ΔBAD,
AB = BA (Common)
∠ABC = ∠BAD = 90°
BC = AD (Given)
ΔABC ≅ ΔBAD [SAS congruency]
Thus,
AC = BD [CPCT]
diagonals are equal.
Now,
In ΔAOB and ΔCOD,
∠BAO = ∠DCO (Alternate interior angles)
∠AOB = ∠COD (Vertically opposite)
AB = CD (Given)
, ΔAOB ≅ ΔCOD [AAS congruency]
Thus,
AO = CO [CPCT].
, Diagonal bisect each other.
Now,
In ΔAOB and ΔCOB,
OB = OB (Given)
AO = CO (diagonals are bisected)
AB = CB (Sides of the square)
, ΔAOB ≅ ΔCOB [SSS congruency]
also, ∠AOB = ∠COB
∠AOB+∠COB = 180° (Linear pair)
Thus, ∠AOB = ∠COB = 90°
, Diagonals bisect each other at right angles
Let ABCD be a square and its diagonals AC and BD intersect each other at O.
To show that,
AC = BD
AO = OC
and ∠AOB = 90°
Proof,
In ΔABC and ΔBAD,
AB = BA (Common)
∠ABC = ∠BAD = 90°
BC = AD (Given)
ΔABC ≅ ΔBAD [SAS congruency]
Thus,
AC = BD [CPCT]
diagonals are equal.
Now,
In ΔAOB and ΔCOD,
∠BAO = ∠DCO (Alternate interior angles)
∠AOB = ∠COD (Vertically opposite)
AB = CD (Given)
, ΔAOB ≅ ΔCOD [AAS congruency]
Thus,
AO = CO [CPCT].
, Diagonal bisect each other.
Now,
In ΔAOB and ΔCOB,
OB = OB (Given)
AO = CO (diagonals are bisected)
AB = CB (Sides of the square)
, ΔAOB ≅ ΔCOB [SSS congruency]
also, ∠AOB = ∠COB
∠AOB+∠COB = 180° (Linear pair)
Thus, ∠AOB = ∠COB = 90°
, Diagonals bisect each other at right angles
Let ABCD be a square and its diagonals AC and BD intersect each other at O.
To show that,
AC = BD
AO = OC
and ∠AOB = 90°
Proof,
In ΔABC and ΔBAD,
AB = BA (Common)
∠ABC = ∠BAD = 90°
BC = AD (Given)
ΔABC ≅ ΔBAD [SAS congruency]
Thus,
AC = BD [CPCT]
diagonals are equal.
Now,
In ΔAOB and ΔCOD,
∠BAO = ∠DCO (Alternate interior angles)
∠AOB = ∠COD (Vertically opposite)
AB = CD (Given)
, ΔAOB ≅ ΔCOD [AAS congruency]
Thus,
AO = CO [CPCT].
, Diagonal bisect each other.
Now,
In ΔAOB and ΔCOB,
OB = OB (Given)
AO = CO (diagonals are bisected)
AB = CB (Sides of the square)
, ΔAOB ≅ ΔCOB [SSS congruency]
also, ∠AOB = ∠COB
∠AOB+∠COB = 180° (Linear pair)
Thus, ∠AOB = ∠COB = 90°
, Diagonals bisect each other at right angles