Given ac is perpendicular to cb and cx is perpendicular to ab, prove angle 2 is congruent to angle 3.

Answer:
Step-by-step explanation:
Given in the question,
AC ⊥ BC and CX ⊥ AB
m∠ACB = 90°
m∠ACX = 90° - m∠3
Now by applying the property of interior angle of triangle,
m∠CAX + m∠CXA + m∠ACX = 180°
m∠2 + 90° + (90°- m∠3) = 180°
m∠2 + 180° - m∠3 = 180°
m∠2 - m∠3 = 0
m∠2 = m∠3
Therefore, m∠2 ≅ m∠3.