Given: AB ≅ BC , m∠MOC = 135° OM − angle bisector of ∠AOB Prove: ∠ABO ≅ ∠CBO
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Answer:
See explanation
Step-by-step explanation:
1. Angles AOM and MOC are supplementry angles. If m∠MOC = 135°, then
m∠AOM = 180° - 135° = 45°
2. OM − angle bisector of ∠AOB, then
m∠AOM = m∠MOB = 45°
3. Now
m∠BOC = m∠MOC - m∠MOB
m∠BOC = 135° - 45° = 90°
4. Since m∠BOC = 90°, BO is perpendicular to AC.
5. Consider isosceles triangle ABC (because AB ≅ BC). BO is the height drawn to the base, so it is an angle B bisector too, thus
∠ABO ≅ ∠CBO