Question

1.Prove that the sum of measures of all angles of a triangle is 180°​

Answer 1

$\huge\star \: \boxed{\underline{\mathtt{\red{a}\pink{ n}\green{s}\blue{w}\purple{e}\orange{r}}}} \: \star</p><p></p><p></p><p>$

To prove the above property of triangles, draw a line

←→

PQ parallel to the side BC of the given triangle.

Since PQ is a straight line, it can be concluded that:

∠PAB + ∠BAC + ∠QAC = 180° ………(1)

Since PQ||BC and AB, AC are transversals,

Therefore, ∠QAC = ∠ACB (a pair of alternate angle)

Also, ∠PAB = ∠CBA (a pair of alternate angle)

Substituting the value of ∠QAC and∠PAB in equation (1),

∠ACB + ∠BAC + ∠CBA= 180°

Thus, the sum of the interior angles of a triangle is 180°.

Answer 2

Step-by-step explanation:

This is your answer....