Question

prove that in an isosceles triangle the bisector of the vertical angle the altitude from the vertex and the median to the base are coincident​

Answer 1

Step-by-step explanation:

i) Let ABC be a triangle, in which AB = AC.

ii) Consider D as midpoint of BC, thus AD is a median of the triangle.

In triangles ADB & ADC,

AB = AC [Given]

BD = CD [Mid point definition]

AD = AD [Common]

So, ΔADB ≅ ΔADC [SSS Congruence Axiom]

==> <BAD ≅ <CAD [Corresponding parts of congruence triangles are equal]

Thus AD is also a bisector of <BAC

As well <ADB ≅ <ADC

But these two are liner pair standing on the same line segment BC.

So sum of them = 180 deg and hence each of them = 90 deg.

So AD is perpendicular to BC

Thus AD is also an altitude of the triangle ABC.

Thus from the above, AD is a Median; AD is an Altitude and AD is an angle bisector.

iii) Similarly considering AD as an angle bisector of <BAC,

we can prove the ΔADB ≅ ΔADC [SAS Congruence axiom]

and hence as above, AD as median, altitude and angle bisector.

Further,taking AD as altitude, the two triangles are congruent under RHS Axiom.

So the above AD as median, altitude and angle bisector.