In the following figure; AC = CD, AD = BD and ∠C = 58°. Find angle CAB.

In △ ACD,

⇒ AC = CD (Given)

∴ ∠CAD = ∠CDA = x (let) (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠CAD + ∠CDA + ∠ACD = 180°

⇒ x + x + 58° = 180°

⇒ 2x = 180° - 58°

⇒ 2x = 122°

⇒ x = $\dfrac{122°}{2}$ = 61°.

⇒ ∠CAD = ∠CDA = 61°.

Since, CDB is a straight line.

∴ ∠CDA + ∠ADB = 180°

⇒ 61° + ∠ADB = 180°

⇒ ∠ADB = 180° - 61° = 119°.

In △ ADB,

⇒ AD = BD (Given)

∴ ∠DAB = ∠DBA = y (let) (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠ADB + ∠DAB + ∠DBA = 180°

⇒ 119° + y + y = 180°

⇒ 119° + 2y = 180°

⇒ 2y = 180° - 119°

⇒ 2y = 61°

⇒ y = $\dfrac{61°}{2}$ = 30.5°

⇒ ∠DAB = ∠DBA = 30.5°

From figure,

⇒ ∠CAB = ∠CAD + ∠DAB

= x + y

= 61° + 30.5°

= 91.5°

Hence, ∠CAB = 91.5°