Modern Geometry Proof about bisectors.

Question

Problem 1: (here you make use any Proposition, up to and including Proposotion 26)

 

Let triangle ABC be an isosceles triangle with segment AB = segment AC. Let D and E be the midpoints of segment AB and AC respectively. Let F be the point of intersection of CD and BE. 

 

1. Prove that Ray AF is the angle bisector of angle A.

 

2. Probe that Ray AF meets segment BC at a point, say G, and that line AG is the perpendicular bisector of segment BC. 

 

This needs to be a formal proof with all parts proven fully. Both parts 1 and 2 need to be proven in this proof, in order for it to be complete. Thank you! 

Answer 1

These are known theorems.

Their proof will be based on the fact that

In the isoceles triangle, the medians from
the base angles are congruent.


Proof:

(Part 1)
Base angles ABC and ACB are congruent.

From 1, medians BE and CD are congruent.

AB and AC are given to be congruent by virtue of the
isoceles triangle.

AF is congruent to itself by reflexive property.

So triangles AFB and AFC are congruent by SSS.

Therefore Angles BAF and CAF are congruent, by
corresponding parts of congruent triangles CPCTC.

This proves that ray AF is indeed the angle bisector.



(Part 2)

With respect to traingles BAG and CAG:

(1) AB = AC by virtue of the given isoceles triangle

(2) Angles ABG and ACG are congruent base angles

(3) AG is congruent to itself

(4) Angles BAG and CAG are congruent by part 1

(5) This forces angles AGB and AGC to be congruent
since two pairs of corresponding angles
are congruent and the angles add up to 180

So triangles BAG and CAG are completely congruent
since we got all 3 pairs of corresponding angles
and 2 pairs of corresponding sides congruent. You
can quote SAS or ASA.

By CPCTC, angles AGB and AGC are congruent.
They are also supplementary because they lie
on CB. Therefore, they are both right angles.
So AG is perpendicular to BC.

Also by CPCTC, BG = GC.
Therefore AG is in fact the perpendicular bisector of BC.

(end of proof)

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You can also try to use the fact that the medians are divided
into a 2 to 1 ratio, but that was not needed here.