Prove that the bisector of any angle is a scalene triangle cannot be perpendicular to the opposite side.?

Answer 1

It helps to draw this out. Take the Sclene triangle ABC. AC is not equal to AC is not equal to CB.
Now imagine we bisect an angle, say ABC with the line BD.
Now if BD is perpedicular to AC, then ADB and CDB are both 90° and are equal.
And clearly the line BD is equal to itself. And since we bisected the angle ABC then angle ABD must be equal to angle CBD.
This means that triangle DAB and triangle DCB are conguent by Angle Side Angle.
If the triangles are congruent side AB must be equal to CB, and our original triangle must be isosceles.

This result contradicts our initial condition that the triangle is scalene, so the bisector of any angle is a scalene triangle cannot be perpendicular to the opposite side.
QED

Answer 2

that may not undemanding, because of the fact the "2 components and an attitude are equivalent" try for congruent triangles would not shop on with because of the fact the anle isn't the coated attitude. enable the triangle be ABC and the mid area of BC be D think AB isn't equivalent to AC. Then take factors P on AB and Q on AC (produced if needed) so as that AP = AC and AQ = AB The diagonals BC and PQ of the quadrilateral BQCP bisect one yet another (BD = DC with the aid of hypothesis, and PDQ is the "replicate image" of BDC contemplated in line advert - or prepare it with the aid of quite some congruent triangles in case you like) as a result BQCP is a parallelogram as a result BP is parallel to high quality controls yet BP strains alongside AB and high quality controls lies alongside AC, and AB and AC at the instant are not parallel because of the fact they are 2 components of a triangle. So there's a contradiction, as a result AB = AC.