Answer:
AO bisects BAC - Needed
AO is perpendicular to BC - Needed
Point O is the midpoint of BC - Not needed
AO is an altitude - Not needed
Step-by-step explanation:
In ΔAOB,
- <BOA = 90° (since AO is perpendicular to BC
)
- <BAO = [tex]\frac{1}{2}[/tex]<BAC (since AO bisects BAC)
In ΔAOC,
- <COA = 90° (since AO is perpendicular to BC
)
- <CAO = [tex]\frac{1}{2}[/tex]<BAC (since AO bisects BAC)
In ΔAOB and ΔAOC,
- <BOA = <COA (since both are 90°)
- <BAO = CAO (since AO bisects BAC)
- AO=AO
So, we can conclude ΔAOB ≅ ΔAOC (ASA property) and hence the base angles <ABO and < ACO of the isosceles triangle ABC are also congruent.