In the following figure, AB = AC, EC = ED, ∠ABF = 45° and ∠ABC = 70°.

Find the angles represented by letters a, b, c, d, e, f and g.

Given: AB = AC (Isosceles triangle)

EC = ED (Isosceles triangle)

∠ABF = 45°

∠ABC = 70°

Since AB = AC, the angles opposite to equal sides are also equal:

⇒ ∠ABC = ∠ACB

⇒ a = 70°

Using the angle sum property of Δ ABC:

⇒ ∠ABC + ∠ACB + ∠BAC = 180°

⇒ 70° + 70° + b = 180°

⇒ 140° + b = 180°

⇒ b = 180° - 140°

⇒ b = 40°

Using the angle sum property in Δ ABF:

⇒ ∠ABF + ∠AFB + ∠BAF = 180°

⇒ 45° + c + 40° = 180°

⇒ 85° + c = 180°

⇒ c = 180° - 85°

⇒ c = 95°

Since ∠AFB and ∠EFC are Vertically opposite angles,

⇒ ∠EFC = c = 95°

Since ∠BAC = b and ∠BAC is alternate to ∠ACE,

⇒ d = b = 40°

Using the angle sum property in Δ EFC,

⇒ ∠EFC + ∠ECF + ∠FEC = 180°

⇒ 95° + 40° + e = 180°

⇒ 135° + e = 180°

⇒ e = 180° - 135° = 45°

Since ∠ACE and ∠CED are alternate angles,

⇒ g = d = 40°

Since EC = ED, the angles opposite to equal sides are also equal.

⇒ ∠ECD = ∠EDC = f

Using the angle sum property in Δ ECD,

⇒ ∠ECD + ∠EDC + ∠DEC = 180°

⇒ f + f + 40° = 180°

⇒ 2f + 40° = 180°

⇒ 2f = 180° - 40°

⇒ 2f = 140°

⇒ f = $\dfrac{140°}{2}$ = 70°

Hence, a = 70°, b = 40°, c = 95°, d = 40°, e = 45°, f = 70° and g = 40°.