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  1. Asked: In:

    ABC is a triangle. Locate a point in the interior of Triangle ABC which is equidistant from all the vertices of Triangle ABC.

    shobha zoon

    Draw perpendicular bisectors of AB, BC, and AC, which intersects each other at G. Point G is equidistant from the three vertices of ΔABC.

    Draw perpendicular bisectors of AB, BC, and AC, which intersects each other at G.
    Point G is equidistant from the three vertices of ΔABC.

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  2. Asked: In:

    In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

    shobha zoon

    Draw the bisectors of ∠A, ∠B and ∠C, which intersect each other at O. Point O is equidistant from the three sides of ΔABC ie. OP = OQ = OR.

    Draw the bisectors of ∠A, ∠B and ∠C, which intersect each other at O. Point O is equidistant from the three sides of ΔABC ie. OP = OQ = OR.

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  3. Asked: In:

    ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

    shobha zoon

    In ΔABC, P is mid-point to AB [∵ Given] Q is mid-point of BC [∵ Given] Hence, PQ ∥ AC and PQ = (1/2)AC ... (1) [∵ Mid-point Theorem] Similarly, in ΔACD, S is mid-point of AD [∵ Given] R is mid-point of CD [∵ Given] Hence, SR ∥ AC and SR = (1/2) AC ...(2) [∵ Mid-Point Theorem] From (1) and (2), we haRead more

    In ΔABC,
    P is mid-point to AB [∵ Given]
    Q is mid-point of BC [∵ Given]
    Hence, PQ ∥ AC and PQ = (1/2)AC … (1) [∵ Mid-point Theorem]

    Similarly, in ΔACD,
    S is mid-point of AD [∵ Given]
    R is mid-point of CD [∵ Given]
    Hence, SR ∥ AC and SR = (1/2) AC …(2) [∵ Mid-Point Theorem]

    From (1) and (2), we have
    PQ ∥ SR …(3) [∵ PQ ∥ AC and SR ∥ AC]
    and PQ = SR …(4) [∵ SR = (1/2)AC and PQ = (1/2)AC]

    Hence, PQRS is a parallelogram.

    Similarly, in ΔBCD,
    Q is mid-point of BC [∵ Given]
    R is mid-point of CD [∵ Given]
    Hence, QR = (1/2) BD …(5) [∵ Mid-point theorem]
    Given that: AC = BD …(6) [∵ Diagonals of a rectangle are equal]
    From (1), (5) and (6), we have
    PQ = QR
    A parallelogram whose adjacent sides are equal, is a rhombus. Hence, PQRS is a rhombus.

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  4. Asked: In:

    Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

    shobha zoon

    In ΔACD, S is mid-point to DA [∵ Given] R is mid-point of DC [∵ Given] Hence, SR ∥ AC and SR = (1/2)AC ... (1) [∵ Mid-point Theorem] In ΔACD, P is mid-point of AB [∵ Given] Q is mid-point of BC [∵ Given] Hence, PQ ∥ AC and PQ = (1/2) AC ...(2) [∵ Mid-Point Theorem] From (1) and (2), we have PQ ∥ SRRead more

    In ΔACD,
    S is mid-point to DA [∵ Given]
    R is mid-point of DC [∵ Given]
    Hence, SR ∥ AC and SR = (1/2)AC … (1) [∵ Mid-point Theorem]

    In ΔACD,
    P is mid-point of AB [∵ Given]
    Q is mid-point of BC [∵ Given]
    Hence, PQ ∥ AC and PQ = (1/2) AC …(2) [∵ Mid-Point Theorem]

    From (1) and (2), we have
    PQ ∥ SR …(3) [∵ PQ ∥ AC and SR ∥ AC]
    and PQ = SR …(4) [∵ SR = (1/2)AC and PQ = (1/2)AC]
    In quadrilateral PQRS,
    PQ ∥ SR and PQ = SR [∵ From (3) and (4)]
    Hence, PQRS is a parallelogram and diagonals of parallelogram bisect each other.
    Therefore, SQ and PR bisects each other.

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  6. Asked: In:

    ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that

    shobha zoon

    (i) In ΔABC, M is mid-point to AB [∵ Given] and DM ∥ BC [∵ Given] Hence, D is mid-point of AC [∵ Converse of Mid-point Theorem] (ii) ∠ADM = ∠ACB [∵ Corresponding angles] ⇒ ∠ADM = 90° [∵ ∠ACB 90°] Hence, MD ⊥ AC (iii) In ΔAMD = ΔCMD, AD = DC [∵ Proved above] ∠ADM = ∠CDM [∵ Each 90°] DM = DM [∵ CommonRead more

    (i) In ΔABC,
    M is mid-point to AB [∵ Given]
    and DM ∥ BC [∵ Given]
    Hence, D is mid-point of AC [∵ Converse of Mid-point Theorem]

    (ii) ∠ADM = ∠ACB [∵ Corresponding angles]
    ⇒ ∠ADM = 90° [∵ ∠ACB 90°]
    Hence, MD ⊥ AC

    (iii) In ΔAMD = ΔCMD,
    AD = DC [∵ Proved above]
    ∠ADM = ∠CDM [∵ Each 90°]
    DM = DM [∵ Common]
    Hence, ΔAMD ΔCMD [∵ SAS Congruency rule]
    AM = CM [∵ CPCT]
    But AM = (1/2) AB [∵ Given]
    Therefore, CM = AM = (1/2) AB

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  7. Asked: In:

    In an isosceles triangle ABC, with AB = AC, the bisectors of Angle B and Angle C intersect each other at O. Join A to O. Show that: (i) OB = OC (ii) AO bisects Angle A

    shobha zoon

    (i) In ABC, AB = AC [∵ Given] Hence, ∠ACB = ∠ACB [∵ Angles opposite to equal sides are equal] ⇒ (1/2)∠ACB = (1/2)∠ABC ⇒ ∠ACO = ∠ABO [∵ OB and OC bisect ∠B and ∠C respectively] In ΔABO and ΔACO, AB = AC [∵ Given] ∠ABO = ∠ACO [∵ Proved above] AO = AO [∵ Common] Hence, Δ ABO ≅ ΔACO [∵ SAS Congruency RuRead more

    (i) In ABC, AB = AC [∵ Given]
    Hence, ∠ACB = ∠ACB [∵ Angles opposite to equal sides are equal]
    ⇒ (1/2)∠ACB = (1/2)∠ABC
    ⇒ ∠ACO = ∠ABO [∵ OB and OC bisect ∠B and ∠C respectively]
    In ΔABO and ΔACO,
    AB = AC [∵ Given]
    ∠ABO = ∠ACO [∵ Proved above]
    AO = AO [∵ Common]
    Hence, Δ ABO ≅ ΔACO [∵ SAS Congruency Rule]
    OB = OC [∵ CPCT]
    (ii) ΔABO ≅ ΔACO [∵ Proved above]
    ∠BAO = ∠CAO [∵ CPCT]
    Hence, OA bisects angle A.

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  8. Asked: In:

    ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠ B and ∠ C.

    shobha zoon

    In ΔABC, AB = AC [∵ Given] ∠B = ∠C [∵ Angles opposite to equal sides are equal] In ΔABC, ∠A + ∠B + ∠C = 180° ⇒ 90° + ∠B + ∠C = 180° [∵ ∠A = 90°] ⇒ 90° + ∠B + ∠B = 180° [∵ ∠C = ∠B] ⇒ 2∠B = 180° - 90° = 90° ⇒ ∠B = (90°/2) = 45° Hence, ∠B = ∠C = 90°

    In ΔABC,
    AB = AC [∵ Given]
    ∠B = ∠C [∵ Angles opposite to equal sides are equal]
    In ΔABC,
    ∠A + ∠B + ∠C = 180°
    ⇒ 90° + ∠B + ∠C = 180° [∵ ∠A = 90°]
    ⇒ 90° + ∠B + ∠B = 180° [∵ ∠C = ∠B]
    ⇒ 2∠B = 180° – 90° = 90°
    ⇒ ∠B = (90°/2) = 45°
    Hence, ∠B = ∠C = 90°

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  9. Asked: In:

    BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

    shobha zoon

    In ΔFBC and ΔECB, ∠BFC = ∠CEB [∵ Each 90°] BC = BC [∵ Common] FC = BE [∵ Given] Hence, ΔFBC ≅ ΔECB [∵ RHS Congruency Rule] ∠FBC = ∠ECB [∵ CPCT] ⇒ AC = AB [∵ Angles opposite to equal Sides are equal] Hence, ΔABC. is an isosceles triangle.

    In ΔFBC and ΔECB,
    ∠BFC = ∠CEB [∵ Each 90°]
    BC = BC [∵ Common]
    FC = BE [∵ Given]
    Hence, ΔFBC ≅ ΔECB [∵ RHS Congruency Rule]
    ∠FBC = ∠ECB [∵ CPCT]
    ⇒ AC = AB [∵ Angles opposite to equal Sides are equal]
    Hence, ΔABC. is an isosceles triangle.

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  11. Asked: In:

    ABC is an isosceles triangle with AB = AC. Draw AP⊥BC to show that Angle B = Angle C.

    shobha zoon

    In ΔABP and ΔACP, ∠APB = ∠APC [∵ Each 90°] AB = AC [∵ Given] AP = AP [∵ Common] Hence, ΔABP ≅ ΔACP [∵ RHS Congruency Rule] ∠B = ∠C [∵ CPCT]

    In ΔABP and ΔACP,
    ∠APB = ∠APC [∵ Each 90°]
    AB = AC [∵ Given]
    AP = AP [∵ Common]
    Hence, ΔABP ≅ ΔACP [∵ RHS Congruency Rule]
    ∠B = ∠C [∵ CPCT]

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  12. Asked: In:

    Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

    shobha zoon

    Given: FB is a line and A is a Point Outside of FB. To Prove: AB is smallest line segment. In ΔABC, ∠B = 90° [∵ Given] Therefore, ∠BAC <90° and ∠ACB ∠ACB [∵ ∠B = 90° and ∠ACB AB [∵ In a triangle, Greater angle has Longer side opposite to it] Similarly, AD > AB, AE > AB and AF > AB, HenceRead more

    Given: FB is a line and A is a Point Outside of FB.
    To Prove: AB is smallest line segment.
    In ΔABC, ∠B = 90° [∵ Given]
    Therefore, ∠BAC <90° and ∠ACB ∠ACB [∵ ∠B = 90° and ∠ACB AB [∵ In a triangle, Greater angle has Longer side opposite to it] Similarly, AD > AB, AE > AB and AF > AB,
    Hence, AB is the Smallest line.

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