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

2 Answers

  • 6 years ago
    Best 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.

  • cidyah
    Lv 7
    6 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)

Still have questions? Get your answers by asking now.