Question

show that the bisector of an angle from vertex of an isosceles triangle bisects the base at right angle

Answer 1

athematics 

 Best Answer

1) By given data, let ABC is an isosceles triangle, in which A is the vertical angle, BC is the base 
and AB = AC; Let the bisector of angle A meets the base BC at D. 

2) We need to prove that BD is perpendicular to BC 

3) Proof: 
Consider the two triangles ABD and ACD, 

AB = AC (Given) 
Angle BAD = Angle CAD (Bisector property; since given AD bisects angle BAC) 
AD = AD (Common side of the two triangles) 

Hence the two triangles ABD and ACD are congruent (By 'SAS' congruence axiom of triangles) 

Therefore angle ADB = angle ADC ------------(1) (Corresponding parts of 
congruence triangles are equal) 

But these two are linear pair that is their sum is 180 degrees --------------(2) 

Hence from (1) and (2), each of the angles ADB and ADC are 90 degrees. 

Therefore AD is perpendicular to BC. (Proved) 

Second Part: 

Suppose now, we consider that AD is not angle bisector as in the previous, but let us consider it as 
Median to the base of the same triangle; 

Here again considering the two triangles ABD and ACD, as previous 

AB = AC (Given) 
AD = AD )Common side) 
BD = CD (Since AD is the median, it divides the base into two equal parts - by definition of the 
median to a side of triangle) 

Hence the two triangles ABD and ACD are congruent (SSS Congruence axion of triangles)

As such again we have, 

angle ADB = angle ADC ------------(1) (Corresponding parts of 
congruence triangles are equal) 

But these two are linear pair that is their sum is 180 degrees --------------(2) 

Hence from (1) and (2), each of the angles ADB and ADC are 90 degrees. 

Therefore AD is perpendicular to BC. (Proved)