CLASS-8
GEOMETRY- ANGLE PROPERTIES OF A TRIANGLE

We should take a triangle ABC.

Draw the line DE parallel to BC through A

Then  ∠ABC = alternate ∠DAB ……………………….(1)

And   ∠ACB = alternate ∠EAC …………………………(2)

Adding (1) & (2), we get –

 ∠ABC + ∠ACB = ∠DAB + ∠EAC

Adding ∠BAC to both sides of the above,

∠BAC + ∠ABC + ∠ACB = ∠DAB + ∠EAC + ∠BAC

                         =  a straight angle =  180⁰

Hence, ∠BAC + ∠ABC + ∠ACB = 180⁰              (Ans.)

Property.2 –

An exterior angle of a triangle is equal to the sum of the two opposite interior angles

Take ∆ ABC. Produce BC to D and through C, draw CE || BA,

Then ∠A = alternate ∠ACE and ∠B = corresponding ∠ECD.

But, ∠ACE + ∠ECD =  ∠ACD

So,  ∠ACD = ∠A + ∠B         (Ans.)