Solution:
In Δ ABC , perpendicular bisector of side AB of ∆ABC intersects side AB at point D and BC at point E. Also, m∠CAB = 82° and m∠C = 68°.
Join A E.
In Δ ABC
∠A+∠B+∠C=180°→→→Angle sum property of triangle.
82°+∠B+68°=180°
∠B= 180°-150°
∠B=30°
In Δ A DE and Δ B DE
AD=B D→→DE is a perpendicular bisector.
∠ADE=∠BDE=90°
DE is common.
Δ A DE ≅ Δ E DB→→[S AS]
∠DEB=∠D A E→→[CPCT]----(1)
In Δ DBE
∠EDB + ∠DBE+∠BED=180°→→Angle sum property of Triangle.
90° +30°+∠BED=180°
∠BED=180°-120°
∠BED=60°
So, ∠BAE=∠BED=60°------[using (1)]
∠CAE=∠CAB - ∠BAE
= 82°-60°
= 22°
