a triangle's median is also an angle bisector of the vertex,prove that the triangle is isosceles triangle.?

2 Answers

  • 7 years ago
    Favorite Answer

    i) Let the triangle be ABC; Ad be the median meeting BC at D.

    Also AD bisects angle BAC.

    ii) By bisector property of an angle of a triangle, 'it divides the opposite base in the same ratio of the other two sides containing the angle.'

    So, AB/AC = BD/CD ---- (1)

    But BD = CD ---- (2) [Since AD being median, it divides BC into two equal parts]

    So from (1) & (2), AB = AC

    Thus the triangle ABC is isosceles.

    ------------ x --------------- x ---------- x ----------- x ------------ x ------------- x------------

    Proof for bisector property:

    i) In the above triangle ABC,through C draw a line parallel to DA to meet BA producedat E.

    ii) Since CE parallel DE, <CAD = <ACE [Alternate interior angles equal] ---- (1)

    Also <BAD = <AEC [Corresponding pair of angles equal] ---- (2)

    But <CAD = <BAD [Bisector divides the angle into two equal parts] --- (3)

    Hence from the above 3 equations, we have, <ACE = <AEC

    So in the triangle AEC, AE = AC ------- (4)

    iii) In triangles BAD and BEC,

    <B = <B [Common]

    <BAD = <BEC [From (2) above]

    So triangle BAD is similar to triangle BEC [AA similarity axiom]

    Hence, BE/BA = BC/BD [Corresponding sides of similar triangles are in proportion]

    Subtracting 1 from both sides, (BE - BA)/BA = (BC - BD)/BD

    ==> AE/AB = CD/BD

    But AE = AC from (4) and BD = CD

    So, AC/AB = CD/BD [Bisector property proved]


    The first contributor 'cidyah' stated it as axiom 'SAS', which I differ; it is only Angle Side Side; but there is no such Angle Side Side congruence axiom exists.

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  • cidyah
    Lv 7
    7 years ago

    Let ABC be a triangle. Let AD be the median. D id the midpoint of BC.

    To prove: AB ≅ AC

    Consider triangles ABD and ACD

    BD ≅ CD --- definition of median

    ∠BAC ≅ ∠CAD --- given

    AD ≅ AD ( reflexive )

    Δ ABD ≅ Δ ACD ( SAS - side angle -side)

    AB ≅ AC ( if two sides and an included angle of one triangle is congruent to corresponding two sides and included angle of another, then the corresponding sides are congruent)

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