1.

Prove that the angle formed by the bisector of interior angle A and the bisector of exterior angle B of a triangle ABC is half of angle C.

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Answer:

Let's draw a ΔABC such that interior BISECTOR of ∠A and exterior bisector of ∠B intersects at E. Also, base AB is produced to D. (please refer to the figure GIVEN below)

∴ ∠CAE = ∠EAB= ½ ∠CAB …… (i)

And,  

∠CBE = ∠EBD= ½ ∠CBD …… (ii)

Let us consider ∠EAB is denoted as “∠x” and ∠EBD is denoted as “∠y”…… (iii)

By exterior angle theorem, we get

∠CBD = ∠A + ∠C

multiplying by ½ throughout the equation

⇒ ½ ∠CBD = ½ ∠A + ½ ∠C

substituting the values from (i), (ii) & (iii)

⇒ ∠y = ∠x + ½ ∠C ……. (iv)

Now, USING the exterior angle theorem in ∆ AEB, we get

∠EBD = ∠x + ∠E

substituting value from (iii)

⇒ ∠y = ∠x + ∠E ….. (V)

Thus, comparing eq. (iv) & (v), we get

∠x + ½ ∠C = ∠x + ∠E

⇒ angle E = ½ ∠C

Hence, it is proved that the angle formed by the bisector of interior angle A and the bisector of exterior angle B of a triangle ABC i.e., angle E is half of angle C.


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