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  1. 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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  2. 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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  3. 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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  4. 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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  6. 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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