ECB intersectboeat a point O prove that217. The side BC of a Δ ABC is

1.	ECB intersectboeat a point O prove that217. The side BC of a Δ ABC is produced on both sides. Prove that the sum of other two exteriorlon an angle of 135 Prove that the triangle isso tormed is greater than Z A by 180
Answer» First you should make a rough sketch of triangle. ∠ABD +∠ABC = 180° ….(i) (Angle on the same line) ∠ACE +∠ACB =180° ……….(ii) Adding (i) and (ii), we get ∠ABD +∠ABC +∠ACE +∠ACB =360° ∠ABD +∠ACE + (∠ABC +∠ACB) =360° …………..(iii) Now, In ΔABC, ∠ABC +∠ACB +∠BAC = 180° ∠ABC +∠ACB = 180°-∠BAC Putting the value of∠ABC +∠ACB in eq.(iii), we get ∠ABD +∠ACE +180°-∠BAC =360° ∠ABD +∠ACE = 360°- 180°+∠BAC ∠ABD +∠ACE = 180°+∠BAC Which proves that sum of the two exterior angles formed is greater than angle A by180° also figure

ECB intersectboeat a point O prove that217. The side BC of a Δ ABC is produced on both sides. Prove that the sum of other two exteriorlon an angle of 135 Prove that the triangle isso tormed is greater than Z A by 180

Answer»

First you should make a rough sketch of triangle.

∠ABD +∠ABC = 180° ….(i) (Angle on the same line)

∠ACE +∠ACB =180° ……….(ii)

Adding (i) and (ii), we get

∠ABD +∠ABC +∠ACE +∠ACB =360°

∠ABD +∠ACE + (∠ABC +∠ACB) =360° …………..(iii)

Now, In ΔABC,

∠ABC +∠ACB +∠BAC = 180°

∠ABC +∠ACB = 180°-∠BAC

Putting the value of∠ABC +∠ACB in eq.(iii), we get

∠ABD +∠ACE +180°-∠BAC =360°

∠ABD +∠ACE = 360°- 180°+∠BAC

∠ABD +∠ACE = 180°+∠BAC

Which proves that sum of the two exterior angles formed is greater than angle A by180°

also figure