Question

Solve the following problem ​

Answer 1

Answer:

$\bold {\pink {\sf\underline{Exterior\: angles \:of\: a \:triangle\: is\: equal\: to\: sum\: of\: all\: opposite\: interior\: angles. }}}$

$\bold {\pink {\sf\underline{Sum\: of\: all\: angles\: of \:a \:triangle\: is= 180°}}}$

Step-by-step explanation:

$\huge {\red{\underline{SOLUTION ⟹}}}$

$\bold {\underline {Given:}}$

• ΔABC
• BC produced to form BC ray and CE is parallel to AB.

$\bold{\underline {To\:Prove:}}$

a). ∠ACD = ∠A + ∠B

b). ∠A + ∠B = ∠ACD = 180°

$\bold {\underline {Proof:}}$

a). CE //AB

⟹ ∠A = ∠ACE [Alternate Interior Angles of parallel lines are equal]........(1)

⟹∠B = ∠ECD [Corresponding angles of parallel lines are equal]............. (2)

Adding equation (1) and (2), we get

⟹ ∠A + ∠B = ∠ACE + ∠ECD

⟹ ∠A + ∠B = ∠ACD [from figure]

Hence, proved exterior angles of a triangle is equal to sum of two opposte interior angles.

b). ∠A + ∠B = ∠ACD [ from above part]......(3)

⟹ ∠ACB + ∠ACD = 180° [ Linear pair]

⟹ ∠ACD = 180° - ∠ACB

Put the value of ∠ACD in equation (3),we get

⟹ ∠A + ∠B = 180° - ∠ACB

⟹ ∠A + ∠B + ∠ACB = 180°

Hence, proved that sum all angels of a triangle is 180°.

$\Large\boxed{PROVED}$

#Be_Brainly

Answer 2

$\bf{\underline{\underline {Given:}}}$

ΔABC

BC produced to form BC ray and CE is parallel to AB.

$\bf{\underline{\underline {To\:Prove:}}}$

a). ∠ACD = ∠A + ∠B

b). ∠A + ∠B = ∠ACD = 180°

$\bf{\underline{\underline{Proof:}}}$

a). CE //AB

⟹ ∠A = ∠ACE [Alternate Interior Angles of parallel lines are equal]........(1)

⟹∠B = ∠ECD [Corresponding angles of parallel lines are equal]............. (2)

Adding equation (1) and (2), we get

⟹ ∠A + ∠B = ∠ACE + ∠ECD

⟹ ∠A + ∠B = ∠ACD [from figure]

b). ∠A + ∠B = ∠ACD [ from above part]......(3)

⟹ ∠ACB + ∠ACD = 180° [ Linear pair]

⟹ ∠ACD = 180° - ∠ACB

Put the value of ∠ACD in equation (3),we get

⟹ ∠A + ∠B = 180° - ∠ACB

⟹ ∠A + ∠B + ∠ACB = 180°

$\bf{All \:angels\:of \:a\: triangle\: is\: 180°.}$