Question

Prove that the line joining the point of intersection of two angular bisectors of the base angles of an isosceles triangles to the vertex bise

Answer 1

Answer:

Bisekt.svg

Let D be a point on the line BC, not equal to B or C and such that AD is not an altitude of triangle ABC.

Let B1 be the base (foot) of the altitude in the triangle ABD through B and let C1 be the base of the altitude in the triangle ACD through C. Then, if D is strictly between B and C, one and only one of B1 or C1 lies inside triangle ABC and it can be assumed without loss of generality that B1 does. This case is depicted in the adjacent diagram. If D lies outside of segment BC, then neither B1 nor C1 lies inside the triangle.

∠ DB1B and ∠ DC1C are right angles, while the angles ∠ B1DB and ∠ C1DC are congruent if D lies on the segment BC (that is, between B and C) and they are identical in the other cases being considered, so the triangles DB1B and DC1C are similar (AAA), which implies that

{\displaystyle {\frac {|BD|}{|CD|}}={\frac {|BB_{1}|}{|CC_{1}|}}={\frac {|AB|\sin \angle BAD}{|AC|\sin \angle CAD}}.}{{\frac {|BD|}{|CD|}}}={{\frac {|BB_{1}|}{|CC_{1}|}}}={\frac {|AB|\sin \angle BAD}{|AC|\sin \angle CAD}}.

If D is the foot of an altitude, then,

{\displaystyle {\frac {|BD|}{|AB|}}=\sin \angle \ BAD{\text{ and }}{\frac {|CD|}{|AC|}}=\sin \angle \ DAC,}{\frac {|BD|}{|AB|}}=\sin \angle \ BAD{\text{ and }}{\frac {|CD|}{|AC|}}=\sin \angle \ DAC,

and the generalized form follows.