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  1. Asked: In:

    Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

    kamal shahu

    Given: ABCD is a quadrilateral in which AO = CO, BO = DO and ∠COD = 90°. To prove: ABCD is a rhombus. Sulution: In ΔAOB and ΔAOD, BO = DO [∵ Given] ∠AOB = ∠AOD [∵ Each 90°] AO = AO [∵ Common] Hence, ΔAOB ≅ ΔAOD [∵ SAS congruency rule] AB = AD [∵ CPCT] Similarly, AB = BC and BC = CD Now, all the fourRead more

    Given: ABCD is a quadrilateral in which AO = CO, BO = DO and ∠COD = 90°.
    To prove: ABCD is a rhombus.
    Sulution: In ΔAOB and ΔAOD,
    BO = DO [∵ Given]
    ∠AOB = ∠AOD [∵ Each 90°]
    AO = AO [∵ Common]
    Hence, ΔAOB ≅ ΔAOD [∵ SAS congruency rule]
    AB = AD [∵ CPCT]
    Similarly, AB = BC and BC = CD
    Now, all the four sides of Quadrilateral ABCD are equal.
    Hence, ABCD is a rhombus.

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  2. Asked: In:

    Show that the diagonals of a square are equal and bisect each other at right angles.

    kamal shahu

    Given: ABCD is a square. To prove: AC = BD, AO = CO, BO = DO and ∠COD = 90°. Solution: ΔBAD and ΔABC, AD = BC [∵ Opposite sides of a square] ∠BAD = ∠ABC [∵ Each 90°] AB = AB [∵ Common] Hence, ΔBAD ≅ ΔABC [∵ SAS Congruency rule] BD = AC [∵ CPCT] In ΔAOB and ΔCOD, ∠OAB = ∠OCD [∵ Alternate angles] AB =Read more

    Given: ABCD is a square.
    To prove: AC = BD, AO = CO, BO = DO and ∠COD = 90°.
    Solution: ΔBAD and ΔABC,
    AD = BC [∵ Opposite sides of a square]
    ∠BAD = ∠ABC [∵ Each 90°]
    AB = AB [∵ Common]
    Hence, ΔBAD ≅ ΔABC [∵ SAS Congruency rule]
    BD = AC [∵ CPCT]
    In ΔAOB and ΔCOD,
    ∠OAB = ∠OCD [∵ Alternate angles]
    AB = CD [∵ Opposite sides of a square]
    ∠OBA = ∠ODC [∵ Alternate angles]
    Hence, ΔBAD ≅ ΔABC [∵ ASA Congruency rule]
    AO = OC, BO = OD, [∵ CPCT]
    In ΔAOB and ΔAOD.
    OB = OD [∵ Proved above]
    AB = AD [∵ Sides of a square]
    OA = OA [∵ Common]
    Hence, ΔBAD ≅ ΔABC [∵ SSS Congruency rule]
    ∠AOB = ∠AOD [∵ CPCT]
    But. ∠AOB + ∠AOD = 180° [∵ Linear pair]
    ⇒ 2∠AOB = 180° [∵ ∠AOD = ∠AOB]
    ⇒ ∠AOB = (180/2) = 90°
    Hence, the diagonals of a square are equal and bisect each at right angles.

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  3. Asked: In:

    Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

    kamal shahu

    Given: ABCD is a quadrilateral such that AC = BD, AO = CO, BO = DO and ∠COD = 90°. To prove: ABCD is a square. Solution: If the diagonals of a quadrilateral bisects each other at right angle, it is a rhombus . Hence, AB = BC = CD = DA In ΔBAD and ΔABC, AD = BC [∵ Proved above] BD = AC [∵ Given] AB =Read more

    Given: ABCD is a quadrilateral such that AC = BD, AO = CO, BO = DO and ∠COD = 90°.
    To prove: ABCD is a square.
    Solution: If the diagonals of a quadrilateral bisects each other at right angle, it is a rhombus .
    Hence, AB = BC = CD = DA
    In ΔBAD and ΔABC,
    AD = BC [∵ Proved above]
    BD = AC [∵ Given]
    AB = AB [∵ Common]
    Hence, ΔBAD ΔABC [∵ SSS Congruency rule]
    ∠BAD = ∠ABC [∵ CPCT]
    But, ∠BAD + ∠ABC = 180° [∵ Co-interior angles]
    ⇒ 2∠ABC = 180° [∵ ∠BAD = ∠ABC]
    ⇒ ∠ABC = (180°/2) = 90°
    Hence, if the diagonals of a quadrilateral are equal and bisect each other at right angles, than it is a square.

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  4. Asked: In:

    If the diagonals of a parallelogram are equal, then show that it is a rectangle.

    kamal shahu

    Given: ABCD is a parallelogram with AC = BD. To Prove: ABCD is a rectangle. Solution: In ΔABC and Δ BAD, BC = AD [∵ Opposite sides of a parallelogram are equal] AC = BD [∵ Given] AB = AB [∵ Common] Hence, ΔABC ≅ ΔBAD [∵ SSS Congruency rule] ∠ABC = ∠BAD [∵ CPCT] But, ∠ABC + ∠BAD = 180° [∵ Co-interiorRead more

    Given: ABCD is a parallelogram with AC = BD.
    To Prove: ABCD is a rectangle.
    Solution: In ΔABC and Δ BAD,
    BC = AD [∵ Opposite sides of a parallelogram are equal]
    AC = BD [∵ Given]
    AB = AB [∵ Common]
    Hence, ΔABC ≅ ΔBAD [∵ SSS Congruency rule]
    ∠ABC = ∠BAD [∵ CPCT]
    But, ∠ABC + ∠BAD = 180° [∵ Co-interior angles]
    ⇒ 2∠BAD = 180° [∵ ∠ABC = ∠BAD]
    ⇒ ∠BAD = 180°/2= 90°
    A parallelogram with one of its is 90° is a rectangle. Hence, ABCD is a rectangle.

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  6. Asked: In:

    Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

    kamal shahu

    Triangles ABD and ABC are on the same base AB and between same parallels, AB∥ CD. Hence, ar (ABD) = ar(ABC) [∵ Triangles on the same base (or equal bases) and between the same parallels are equal in area.] Subtracting ar (ABO) from both of sides ar(ABD) - ar(ABO) = ar(ABC) - ar(ABO) ⇒ ar(AOD) = ar(BRead more

    Triangles ABD and ABC are on the same base AB and between same parallels, AB∥ CD.
    Hence, ar (ABD) = ar(ABC)
    [∵ Triangles on the same base (or equal bases) and between the same parallels are equal in area.]
    Subtracting ar (ABO) from both of sides
    ar(ABD) – ar(ABO) = ar(ABC) – ar(ABO)
    ⇒ ar(AOD) = ar(BOC)

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