Question

IN the given figure, the bisectorsof angleABC and angleBCA intersect each other at O. prove that angle BOC =90degree+1/2angleA

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

In ∆ ABC

Bisector of ∠ABC and ∠ACB bisects at O.

Since, OB bisects ∠ABC

⇛ ∠ABO = ∠OBC = x say ----(1)

Again, OC bisects ∠ACB

⇛ ∠OCA = ∠OCB = y say -----(2)

Now, Let we assume that ∠ABC = w and ∠BOC = z

Now, In ∆ BOC

We know, Sum of all interior angles of a triangle is supplementary.

⇛ ∠OBC + ∠OCB + ∠BOC = 180°.

⇛ x + y + z = 180°

⇛ x + y = 180° - z -----(3)

Now, In ∆ ABC

We know, Sum of all interior angles of a triangle is supplementary.

⇛ ∠ABC + ∠ACB + ∠BAC = 180°

⇛ 2x + 2y + w = 180°

⇛ 2( x + y ) + w = 180°

⇛ 2( 180° - z) + w = 180° [ using (1) ]

⇛ 360° - 2z + w = 180°

⇛ 2z = 360° - 180° + w

⇛ 2z = 180° + w

$\bf\implies \:z = 90 \degree \: + \: \dfrac{1}{2}w$

$\bf\implies \: \angle \: BOC = 90 \degree \: + \: \dfrac{1}{2} \angle \: A$

Hence, Proved

Properties of a triangle

Angle Sum Property of triangle :- The sum of all interior angles of a triangle is supplementary. 

The sum of two sides of a triangle is always greater than the third side.

The side opposite to the largest angle of a triangle is the largest side.

The angle opposite to greatest side is always larger.

Exterior angle Property of the triangle :- Exterior angle of a triangle is equal to the sum of its interior opposite angles.

Based on the angle measurement, there are three types of triangles:

Acute Angled Triangle : A triangle having all three angles less than 90° is an acute angle triangle.

Right-Angled Triangle : A triangle that has one angle 90° is a right-angle triangle.

Obtuse Angled Triangle : A triangle having one angle more than 90° is an obtuse angle triangle.