Question

I request you please help me Q.16​

Answer 1

Step-by-step explanation:

Given: △ABC in which AB,BC and CA are produced to points D,E and F.

To prove: ∠DCA+∠FAE+∠FBD=180

∘

Proof :

From the figure we know that

∠DCA=∠A+∠B.(1)

∠FAE=∠B+∠C.(2)

∠FBD=∠A+∠C.(3)

By adding equation (1),(2) and (3) we get

∠DCA+∠FAE+∠FBD=∠A+∠B+∠B+∠C+∠A+∠C

So we get

∠DCA+∠FAE+∠FBD=2∠A+2∠B+2∠C

Now by taking out 2 as common

∠DCA+∠FAE+∠FBD=2(∠A+∠B+∠C)

We know that the sum of all the angles in a triangle is 180

∘

.

So we get

∠DCA+∠FAE+∠FBD=2(180

∘

)

∠DCA+∠FAE+∠FBD=360

∘

HOPE IT HELPS MATE

Answer 2

If the sides of a triangle are produced in order, prove that the sum of the exterior angles so formed is equal to four right angles.

Given: △ABC in which AB,BC and CA are produced to points D,E and F.

To prove: ∠DCA+∠FAE+∠FBD=180∘

Proof :

From the figure we know that

∠DCA=∠A+∠B.(1)

∠FAE=∠B+∠C.(2)

∠FBD=∠A+∠C.(3)

By adding equation (1),(2) and (3) we get

∠DCA+∠FAE+∠FBD=∠A+∠B+∠B+∠C+∠A+∠C

So we get

∠DCA+∠FAE+∠FBD=2∠A+2∠B+2∠C

Now by taking out 2 as common

∠DCA+∠FAE+∠FBD=2(∠A+∠B+∠C)

We know that the sum of all the angles in a triangle is 180

So we get,

∠DCA+∠FAE+∠FBD=2(180)

∠DCA+∠FAE+∠FBD=360

Therefore, it is proved.

.