In the following figure, ABCD is a rhombus and DCFE is a square.

If ∠ABC = 56°, find :

(i) ∠DAE

(ii) ∠FEA

(iii) ∠EAC

(iv) ∠AEC

(i) In rhombus ABCD,

⇒ ∠CDA = ∠CBA = 56° (Opposite angles of rhombus are equal.)

In square DCFE,

⇒ ∠CDE = 90° (Each interior angle of a square equals to 90°)

From figure,

⇒ AD = CD (Each side of rhombus are equal) ………(1)

⇒ CD = ED (Each side of square are equal) ………(2)

From equation (1) and (2), we get :

⇒ AD = ED.

In △ ADE,

⇒ AD = ED (Proved above)

⇒ ∠AED = ∠DAE

By angle sum property of triangle,

⇒ ∠DAE + ∠AED + ∠ADE = 180°

⇒ ∠DAE + ∠DAE + (∠CDA + ∠CDE) = 180°

⇒ 2∠DAE + (56° + 90°) = 180°

⇒ 2∠DAE + 146° = 180°

⇒ 2∠DAE = 180° - 146°

⇒ 2∠DAE = 34°

⇒ ∠DAE = $\dfrac{34°}{2}$ = 17°.

Hence, ∠DAE = 17°.

(ii) From figure,

⇒ ∠FED = 90° (Each interior angle of square equals 90°)

⇒ ∠AED = ∠DAE = 17°.

⇒ ∠FEA = ∠FED - ∠AED

⇒ ∠FEA = 90° - 17° = 73°.

Hence, ∠FEA = 73°.

(iii) In rhombus ABCD,

By angle sum property,

⇒ ∠ABC + ∠BCD + ∠CDA + ∠DAB = 360°

⇒ ∠ABC + ∠DAB + ∠CDA + ∠DAB = 360° (∠BCD = ∠DAB, as opposite angles of rhombus are equal)

⇒ 56° + 2∠DAB + 56° = 360°

⇒ 112° + 2∠DAB = 360°

⇒ 2∠DAB = 360° - 112°

⇒ 2∠DAB = 248°

⇒ ∠DAB = $\dfrac{248°}{2}$ = 124°.

From figure,

⇒ ∠DAC = $\dfrac{∠DAB}{2}$ (As, diagonals of rhombus bisect interior angles)

⇒ ∠DAC = $\dfrac{124°}{2}$ = 62°.

From figure,

⇒ ∠EAC = ∠DAC - ∠DAE = 62° - 17° = 45°.

Hence, ∠EAC = 45°.

(iv) Join EC.

From figure,

⇒ ∠DEC = $\dfrac{∠DEF}{2}$ (As, diagonals of square bisect interior angles)

⇒ ∠DEC = $\dfrac{90°}{2}$ = 45°.

From figure,

⇒ ∠AEC = ∠DEC - ∠DEA = 90° - 17° = 73°.

Hence, ∠AEC = 73°.