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

Relevance

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

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]

EDIT:

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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• 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

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