Question

15. The bisector of ∠B and ∠C intersect each other at a point O. Prove that ∠BOC = 90° + ½ ∠A​

Answer 1

Given :

➤ The bisector of ∠B and ∠C intersect each other at a point O.

To Prove :

➤ ∠BOC = 90° + ½ ∠A

Proof :

For solving this question let's first know about Traingle.

A traingle is also a polygon.

• A traingle has three sides.
• A traingle has three vertices.
• A traingle has three angles.
• The sum of the three interior angles of a triangle is 180°.

Let's solve this question...

As we know that

The sum of the three interior angles of a triangle = 180°

So ,

In ∆ABC

$\tt{: \implies \angle A + \angle B + \angle C = {180}^{ \circ} }$

$\tt{: \implies \dfrac{1}{2} \angle A + \dfrac{1}{2} \angle B +\dfrac{1}{2} \angle C = \cancel{\dfrac{180}{2} }}$

$\tt{: \implies \dfrac{1}{2} \angle A + \angle OBC + \angle OCB = {90}^{ \circ} }$

$\tt{: \implies \angle OBC + \angle OCB = {90}^{ \circ} - \dfrac{1}{2} \: \: - - - \bf{ \big \lgroup 1 \big\rgroup}}$

In ∆DBC

$\tt{: \implies (\angle OBC + \angle OCB) + \angle BOC = {180}^{ \circ} }$

$\tt{: \implies {90}^{ \circ} - \dfrac{1}{2} \angle A + \angle BOC = {180}^{ \circ} \bf \bigg \lgroup Using \: 1 \bigg \rgroup }$

$\tt{: \implies \angle BOC = {180}^{ \circ} - {90}^{ \circ} + \dfrac{1}{2} \angle A}$

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

Hence, Proved !!

Answer 2

Answer:

➤ The bisector of ∠B and ∠C intersect each other at a point O.

To Prove :

➤ ∠BOC = 90° + ½ ∠A

Proof :

For solving this question let's first know about Traingle.

A traingle is also a polygon.

A traingle has three sides.

A traingle has three vertices.

A traingle has three angles.

The sum of the three interior angles of a triangle is 180°.

Let's solve this question...

As we know that

The sum of the three interior angles of a triangle = 180°

So ,

In ∆ABC

[tex] \tt{: \implies \angle A + \angle B + \angle C = {180}^{ \circ} }:⟹∠A+∠B+∠C=180

∘

\tt{: \implies \dfrac{1}{2} \angle A + \dfrac{1}{2} \angle B +\dfrac{1}{2} \angle C = \cancel{\dfrac{180}{2} }}:⟹

2

1

∠A+

2

1

∠B+

2

1

∠C=

2

180

\tt{: \implies \dfrac{1}{2} \angle A + \angle OBC + \angle OCB = {90}^{ \circ} }:⟹

2

1

∠A+∠OBC+∠OCB=90

∘

\tt{: \implies \angle OBC + \angle OCB = {90}^{ \circ} - \dfrac{1}{2} \: \: - - - \bf{ \big \lgroup 1 \big\rgroup}}:⟹∠OBC+∠OCB=90

∘

−

2

1

−−−

⎩

⎪

⎧

1

⎭

⎪

⎫

In ∆DBC

\tt{: \implies (\angle OBC + \angle OCB) + \angle BOC = {180}^{ \circ} }:⟹(∠OBC+∠OCB)+∠BOC=180

∘

\tt{: \implies {90}^{ \circ} - \dfrac{1}{2} \angle A + \angle BOC = {180}^{ \circ} \bf \bigg \lgroup Using \: 1 \bigg \rgroup }:⟹90

∘

−

2

1

∠A+∠BOC=180

∘

⎩

⎪

⎪

⎪

⎧

Using1

⎭

⎪

⎪

⎪

⎫

\tt{: \implies \angle BOC = {180}^{ \circ} - {90}^{ \circ} + \dfrac{1}{2} \angle A}:⟹∠BOC=180

∘

−90

∘

+

2

1

∠A

\bf{: \implies \angle BOC = {90}^{ \circ} + \dfrac{1}{2} \angle A}:⟹∠BOC=90

∘

+

2

1

∠A

Hence, Proved !!