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.
Answer:
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