Question

$\green{QUESTION}$
If the sides of a triangle are produced in an order , show that the sum of the exterior angles so formed is 360°.

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Answer 1

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

To prove: ∠DCA + ∠FAE + ∠FBD = 180o 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 180degree.

So we get ∠DCA + ∠FAE + ∠FBD = 2 (180o)

∠DCA + ∠FAE + ∠FBD = 360degree

Therefore, it is proved.Read more on

Answer 2

Answer:

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.

Step-by-step explanation:

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