a triangle's median is also an angle bisector of the vertex,prove that the triangle is isosceles triangle.?
- LearnerLv 76 years agoBest 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.
- cidyahLv 76 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)