The perpendicular bisector of side ab of ∆abc intersects side ab at point d and bc at point e. if m∠cab = 82° and m∠c = 68°, find m∠cae.

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°

55 degrees because 55+55=110 degrees