Given: ab ≅ bc , m∠moc = 135° om − angle bisector of ∠aob prove: ∠abo ≅ ∠cbo

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

to find the perimeter of the rectangle this is the expression:

x-9+x-9+3x-4+3x-4=138

if we simplify it:

8x-26=138

remember, 138 is what it all adds up to

the length is 11.5cm

the width is 57.5cm

the area is 661.25cm^2

the answers might not be correct use at own risk

no idea

step-by-step explanation: