1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. Let the angles of the quadrilateral be 3x, 5x, 9x and 13x. ∵ Sum of all the angles of quadrilateral = 360º ∴ 3x + 5x + 9x + 13x = 360º ⇒ 30x = 360º ⇒ x=(360o/30)= 12° ∴ 3x = 3 x 12° = 36º 5x = 5 x 12° = 60º 9x = 9 x 12° = 108º 13x = 13 x 12° = 156º ⇒ The required angles of the quadrilateral are 36º, 60º, 108º and 156º. 2. If the diagonals of a parallelogram are equal, then show that it is a rectangle. A parallelogram ABCD such that AC = BD In ΔABC and ΔDCB, AC = DB [Given] AB = DC [Opposite sides of a parallelogram] BC = CB [Common] ΔABC ≌ ΔDCB [SSS criteria] ∴ Their corresponding parts are equal. ⇒ ∠ABC = ∠DCB …(1) ∵ AB || DC and BC is a transversal. [∵ ABCD is a parallelogram] ∴ ∠ABC + ∠DCB = 180º [Interior opposite angles are supplementary] …(2) From (1) and (2), we have ∠ABC = ∠DCB = 90º i.e. ABCD is parallelogram having an angle equal to 90º. ∴ ABCD is a rectangle. 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. We have a quadrilateral ABCD such that the diagonals AC and BD bisect each other at right angles at O. ∴ In ΔAOB and ΔAOD, we have AO = AO [Common] OB = OD [Given that O in the mid-point of BD] ∠AOB = ∠AOD [Each = 90°] ∴ ΔAOB ≌ ΔAOD [SAS criteria] ⇒ Their corresponding parts are equal. ∴ AB = AD …(1) Similarly, AB = BC …(2) BC = CD …(3) CD = AD …(4) ∴ From (1), (2), (3) and (4), we have AB = BC = CD = DA Thus, the quadritateral ABCD is a rhombus. 4. Show that the diagonals of a square are equal and bisect each other at right angles. We have a square ABCD such that its diagonals AC and BD intersect at O. (i) To prove that the diagonals are equal, i.e. AC = BD In ΔABC and ΔBAD, we have AB = BA [Common] BC = AD [Opposite sides of the square ABCD] ∠ABC = ∠BAD [All angles of a square are equal to 90º] ∴ DABC ≌ DBAD [SAS criteria] ⇒ Their corresponding parts are equal. ⇒ AC = BD …(1) (ii) To prove that ‘O’ is the mid-point of AC and BD. ∵ AD || BC and AC is a transversal. [∵ Opposite sides of a square are parallel] ∴ ∠1 = ∠3 [Interior alternate angles] Similarly, ∠2= ∠4 [Interior alternate angles] Now, in ΔOAD and ΔOCB, we have AD = CB [Opposite sides of the square ABCD] ∠1= ∠3 [Proved] ∠2= ∠4 [Proved] ∴ ΔOAD ≌ ΔOCB [ASA criteria] ∴ Their corresponding parts are equal. ⇒ OA = OC and OD = OB ⇒ O is the mid-point of AC and BD, i.e. the diagonals AC and BD bisect each other at O. …(2) (iii) To prove that AC ⊥ BD. In ΔOBA and ΔODA, we have OB = OD [Proved] BA = DA [Opposite sides of the square] OA = OA [Common] ∴ ΔOBA ≌ ΔODA [SSS criteria] ⇒ Their corresponding parts are equal. ⇒ ∠AOB = ∠AOD But ∠AOB and ∠AOD form a linear pair. ∴ ∠AOB + ∠AOD = 180º ⇒ ∠AOB = ∠AOD = 90º ⇒ AC ⊥ BD …(3) From (1), (2) and (3), we get AC and BD are equal and bisect each other at right angles. 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. We have a quadrilateral ABCD such that ‘O’ is the mid-point of AC and BD. Also AC ⊥ BD. Now, in ΔAOD and ΔAOB, we have AO = AO [Common] OD = OB [∵ O is the mid-point of BD] ∠AOD = ∠AOB [Each = 90º] ∴ ΔAOD ≌ ΔAOB [SAS criteria] ∴ Their corresponding parts are equal. ⇒ AD = AB …(1) Similarly, we have AB = BC …(2) BC = CD …(3) CD = DA …(4) From (1), (2), (3) and (4) we have: AB = BC = CD = DA ∴ Quadrilateral ABCD is having all sides equal. In ΔAOD and ΔCOB, we have AO = CO [Given] OD = OB [Given] ∠AOD = ∠COB [Vertically opposite angles] ∴ ΔAOD ≌ ΔCOB ⇒ Their corresponding parts are equal. ⇒ ∠1= ∠2 But, they form a pair of interior alternate angles. ∴ AD || BC Similarly, AB || DC ∴ ABCD is a parallelogram. ∵ Parallelogram having all of its sides equal is a rhombus. ∴ ABCD is a rhombus. Now, in ΔABC and ΔBAD, we have AC = BD [Given] BC = AD [Proved] AB = BA [Common] ∴ ΔABC ≌ ΔBAD [SSS criteria] ⇒ Their corresponding angles are equal. ∴ ∠ABC = ∠BAD Since, AD || BC and AB is a transversal. ∴ ∠ABC + ∠BAD = 180º [Interior opposite angles are supplementary] i.e. The rhombus ABCD is having one angle equal to 90º. Thus, ABCD is a square. 6. Diagonal AC of a parallelogram ABCD bisects ∠ A (see figure). Show that (i) it bisects ∠ C also, (ii) ABCD is a rhombus. Solution: We have a parallelogram ABCD in which diagonal AC bisects ∠A. ⇒ ∠DAC = ∠BAC (i) To prove that AC bisects ∠C. ∵ ABCD is a parallelogram. ∴ AB || DC and AC is a transversal. ∴ ∠1 = ∠3[Alternate interior angles] …(1) Also, BC || AD and AC is a transversal. ∴ ∠2 = ∠4[Alternate interior angles] …(2) But AC bisects ∠A. [Given] ∴ ∠1 = ∠2 …(3) From (1), (2) and (3), we have ∠3= ∠4 ⇒ AC bisects ∠C. (ii) To prove ABCD is a rhombus. In DABC, we have ∠1= ∠4[∵ ∠1 = ∠2 = ∠4] ⇒ BC = AB [Sides opposite to equal angles are equal] …(4) Similarly, AD = DC …(5) But ABCD is a parallelogram [Given] ∴ AB = DC [Opposite sides of a parallelogram] …(6) From (4), (5) and (6), we have AB = BC = CD = DA Thus, ABCD is a rhombus. 7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D. ABCD is a rhombus. ∴ AB = BC = CD = AD Also, AB || CD an AD || BC Now, AD = CD ⇒ ∠1 = ∠2 …(1) [Angles opposite to equal sides are equal] Also, CD || AB [Opposite sides of the parallelogram] and AC is AC is transversal. ∴ ∠1= ∠3 [Alternate interior angles] …(2) From (1) and (2), we have ∠2= ∠3 and ∠1 = ∠4 ⇒ AC bisects ∠C as well as ∠A. Similarly, we prove that BD bisects ∠B as well as ∠D. 8. ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C. Show that: (i) ABCD is a square (ii) diagonal BD bisects ∠ B as well as ∠ D. Solution: We have a rectangle ABCD such that AC bisects ∠ A as well as ∠C. i.e. ∠ 1= ∠ 4 and ∠ 2 = ∠ 3 ...(1) (i) Since, rectangle is a parallelogram. ∴ ABCD is a parallelogram. ⇒ AB || CD and AC is a transversal. ∴ ∠2= ∠4 [Alternate interior angles] . ..(2) From (1) and (2), we have ∠3= ∠4 ⇒ AB = BC [∵ Sides opposite to equal angles in D ABC are equal.] ∴ AB = BC = CD = AD ⇒ ABCD is a rectangle having all of its sides equal. ∴ ABCD is a square. (ii) Since, ABCD is a square, and diagonals of a square bisect the opposite angles. ∴ BD bisects ∠B as well as ∠D.