Question #304c8

2 Answers
Jan 23, 2018

Yes.
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Explanation:

In the case of isosceles and equilateral triangles, a median bisects any angle at a vertex whose two adjacent sides are equal in length.

An altitude of a triangle is a line segment through a vertex and perpendicular to a line containing the base.

Thus, for an isosceles triangle the median and altitude will be the same for the angle-side pair are from the intersection point of the two sides of equal length.

https://mathopenref.com/triangle.html

Jan 23, 2018

HenceAD is the median, altitude & angular bisector of the isosceles triangle.

Explanation:

enter image source here

Given : It’s an isosceles triangle with sides #AC = AB# & #/_C = /_B#

Let AB be the angular bisector of #/_A#.

Using angular bisector theorem,

#(AC) / (AB) = 1 =(CD) / (CB)#

#:. CD = CB#

i.e. AD is the median of BC.

Triangles ACD & ABD are congruent as

#AC = AB, CD = BD, AD common.

#:. /_(ADC) = /_(ADB =90^0#

Hence AD is also the altitude on CB.

HenceAD is the median, altitude & angular bisector of the isosceles triangle.