Answer: ∠FBD is equal to 90° and the angles of triangle ABC are ∠A = 95°, ∠C = 65°, ∠ABE= 20°

Step-by-step explanation:

Here given ∠FBC=70°

∠EAC=25°

∠FBE=∠AEB=∠AEC=90°

FB is perpendicular to DBC

AE is perpendicular bisector to BC

To find ∠FBD, all the angles of triangle ABC.

According to question,

To find ∠FBD we know that, FB is perpendicular to DBC so ∠FBD is equal to 90°.

Now to find angles of △ABC :-

∠ABE = ∠ABF + ∠ABE

90° = 70° + ∠ABE

∠ABE= 20°

As we know △AEC :-

∠AEC = 90° { Right angle triangle}

∠CAE = 25°

so, ∠C = 180° - (∠AEC + ∠CAE)

∠C = 180° - (90°+ 25°)

∠C = 65°

So angle ∠A = 180° - ( ∠C + ∠ABC)

∠A = 180° - ( ∠C + ∠ABC)

∠A = ∠EAB + ∠CAE

∠EAB = 180° - ( ∠C + ∠ABC) - ∠CAE

∠EAB = 180° - ( 65° + 20° ) - 25°

∠EAB = 70°

SO ∠A = ∠EAB + ∠CAE

∠A = 70° + 25°

∠A = 95°

Therefore ∠FBD is equal to 90 and the angles of triangle ABC are ∠A = 95°, ∠C = 65°, ∠ABE= 20°