Please help sum 12(ii)

Please help sum 12(ii) Prove that each angle of a rectangle is 900. that it is a rectangle, (if) If the angle of a quadrilateral are equal, prove If the diagonals of a rhombus are equal, prove that it is a square. (it') Prove that every diagonal of a rhombus bisects the angles at the vertices. g, ABCD parallelogram If the diagonal AC bisects LA, then prove that: (i) AC bisects Z C (ii) ABCD is a rhombus (iil) AC -L BD. (i) Prove that bisectors of any two adjacent angles of a parallelogram are at right (ii) Prove that bisectors of any two opposite angles of a parallelogram are parallel. (iii) If the diagonals of a quadrilateral are equal and bisect each Other at right angles, then prove that it is a square. (D If A BCD is a rectangle in which the diagonal BD bisects Z B, then show that ABCD is a square. (it) Show that if the diagonals of a quadrilateral are equal and bisect each Other at right angles, then it is a square. p and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O. Hint. Show that AOAP AOCQ. (a) In figure ( I ) given below, ABCD is a parallelogram and X is mid-point Of BC. The line AX produced meets DC produced at Q The parallelogram ABPQ is completed. Prove that (1) the triangles ABX and QCX are congruent. (iO DC QI' (b) In figure (2) given below, points P and Q have been taken on opposite sides AB and CD respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. c Q o B (2) ABCD is a square. A is joined to a point P on BC and D is joined to a point Q on AB. If AP DQ prove that AP and DQ are perpendicular to each other. Hint. AABP DAQ ZBAP = LADQ. But LBAD = 900 ZPAD + ZADQ = 90'. If P and Q are points of trisection of the diagonal BD of a parallelogram ABCD, prove that CQ II AP. Hint. A ABP ACDQ. A transversal cuts two parallel lines at A and B. The two interior angles at A are bisected md so are the two interior angles at B; the four bisectors form a quadrilateral ACBD

Dear Student, 
 

Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O. It is given that the diagonals of ABCD are equal and bisect each other at right angles. Therefore, AC = BD, OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90º. 

Now, to prove ABCD is a square, we have to prove that ABCD is a parallelogram, AB = BC = CD = AD, and one of its interior angles is 90º.

Proof:

In ΔAOB and ΔCOD,

AO = CO (Diagonals bisect each other)

OB = OD (Diagonals bisect each other)

∠AOB = ∠COD (Vertically opposite angles)

∴ ΔAOB ≅ ΔCOD (SAS congruence rule)

∴ AB = CD (By CPCT) ... (1)

And, ∠OAB = ∠OCD (By CPCT)

However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel.

∴ AB || CD ... (2)

From equations (1) and (2), we obtain

ABCD is a parallelogram.

In ΔAOD and ΔCOD,

AO = CO (Diagonals bisect each other)

∠AOD = ∠COD (Given that each is 90º)

OD = OD (Common)

∴ ΔAOD ≅ ΔCOD (SAS congruence rule)

∴ AD = DC ... (3)

However, AD = BC and AB = CD (Opposite sides of parallelogram ABCD)

∴ AB = BC = CD = DA

Therefore, all the sides of quadrilateral ABCD are equal to each other.

In ΔADC and ΔBCD,

AD = BC (Already proved)

AC = BD (Given)

DC = CD (Common)

∴ ΔADC ≅ ΔBCD (SSS Congruence rule)

∴ ∠ADC = ∠BCD (By CPCT)

However, ∠ADC + ∠BCD = 180° (Co-interior angles)

⇒ ∠ADC + ∠ADC = 180°

⇒ 2∠ADC = 180°

⇒ ∠ADC = 90°

One of the interior angles of quadrilateral ABCD is a right angle.

Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angles is 90º. Therefore, ABCD is a square.


Regards

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