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  1. Asked: In:

    ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Figure). Show that these altitudes are equal.

    sanjeev rao

    In ΔABE and ΔACF, ∠AEB = ∠AFC [∵ Each 90°] ∠A = ∠A [∵ Common] AB = AC [∵ Given] Hance, ΔABE ≅ ΔACF [∵ SAS Congruency Rule] BE = CF [∵ CPCT]

    In ΔABE and ΔACF,
    ∠AEB = ∠AFC [∵ Each 90°]
    ∠A = ∠A [∵ Common]
    AB = AC [∵ Given]
    Hance, ΔABE ≅ ΔACF [∵ SAS Congruency Rule]
    BE = CF [∵ CPCT]

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  2. Asked: In:

    In Δ ABC, AD is the perpendicular bisector of BC (see Figure). Show that Δ ABC is an isosceles triangle in which AB = AC.

    sanjeev rao

    In ΔABD and ΔACD, BD = DC [∵ AD bisects BC] ∠ADB = ∠ADC [∵ Each 90°] AD = AD [∵ Common] Hance, ΔABD ≅ ΔACD [∵ SAS Congruency Rule] AB = AC [∵ CPCT] Hence, ABC is an isosceles triangle.

    In ΔABD and ΔACD,
    BD = DC [∵ AD bisects BC]
    ∠ADB = ∠ADC [∵ Each 90°]
    AD = AD [∵ Common]
    Hance, ΔABD ≅ ΔACD [∵ SAS Congruency Rule]
    AB = AC [∵ CPCT]
    Hence, ABC is an isosceles triangle.

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  3. Asked: In:

    ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Figure). Show that (i) Δ ABE ≅ Δ ACF (ii) AB = AC, i.e., ABC is an isosceles triangle.

    sanjeev rao

    (i) In ΔABE and ΔACF, ∠AEB = ∠AFC [∵ Each 90] ∠A = ∠A [∵ Common] BE = CF [∵ Given] Hance, ΔABE ≅ ΔACF [∵ SAS Congruency Rule] (ii) In ΔABE ΔACF [∵ Proved above] AB = AC [∵ CPCT] Hence, ΔABC is an isosceles triangle.

    (i) In ΔABE and ΔACF,
    ∠AEB = ∠AFC [∵ Each 90]
    ∠A = ∠A [∵ Common]
    BE = CF [∵ Given]
    Hance, ΔABE ≅ ΔACF [∵ SAS Congruency Rule]

    (ii) In ΔABE ΔACF [∵ Proved above]
    AB = AC [∵ CPCT]
    Hence, ΔABC is an isosceles triangle.

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  4. Asked: In:

    ABC and DBC are two isosceles triangles on the same base BC (see Figure). Show that ∠ ABD = ∠ ACD.

    sanjeev rao

    In ΔABC, AB = AC [∵ Given] ∠ABC = ∠ACB ...(1) [∵ Angles opposite to equal sides are equal] In ΔDBC, DB = DC [∵ Given] ∠DBC = ∠DCB ...(2) [∵ Angles opposite to equal sides are equal] Adding equation (1) and (2), we get ∠ABC + ∠DBC = ∠ACB + ∠DCB ⇒ ∠ABD = ∠ACD.

    In ΔABC,
    AB = AC [∵ Given]
    ∠ABC = ∠ACB …(1) [∵ Angles opposite to equal sides are equal]
    In ΔDBC,
    DB = DC [∵ Given]
    ∠DBC = ∠DCB …(2) [∵ Angles opposite to equal sides are equal]
    Adding equation (1) and (2), we get
    ∠ABC + ∠DBC = ∠ACB + ∠DCB
    ⇒ ∠ABD = ∠ACD.

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  6. Asked: In:

    ΔABC is an isosceles triangle in which AB = AC.Side BA is produced to D such that AD = AB (see Figure). Show that ∠ BCD is a right angle.

    sanjeev rao

    In ΔACD, AB = AC [∵ Given] ∠ACD = ∠D ...(1) [∵ Angles opposite to equal sides are equal] In ΔABC, AB = AC [∵ Given] ∠B = ∠ACB ...(2) [∵ Angles opposite to equal sides are equal] In ΔDBC, ∠D + ∠B + ∠BCD = 180° ⇒ ∠ACD + ∠ACB + BCD = 180° [∵ From the equation (1) and (2)] ⇒ ∠BCD + ∠BCD = 180° [∵ ∠ACD +Read more

    In ΔACD,
    AB = AC [∵ Given]
    ∠ACD = ∠D …(1) [∵ Angles opposite to equal sides are equal]
    In ΔABC,
    AB = AC [∵ Given]
    ∠B = ∠ACB …(2) [∵ Angles opposite to equal sides are equal]
    In ΔDBC,
    ∠D + ∠B + ∠BCD = 180°
    ⇒ ∠ACD + ∠ACB + BCD = 180° [∵ From the equation (1) and (2)]
    ⇒ ∠BCD + ∠BCD = 180° [∵ ∠ACD + ∠ACB = ∠BCD]
    ⇒ 2∠BCD = 180°
    ⇒ ∠BCD = (180°/2) = 90°
    Hence, ∠BCD is a is a right angle.

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