Question

In triangle ABC (fig.(b)) , internal bisectors of <B and <C meet at p and external bisectors of these angles meet at Q . prove that <BPC + <BQC = 180°
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Answer 1

Step-by-step explanation:

Answer

∠ABC+ext.∠∠ABC=180°

(Angles on a straight line)21

(∠ABC+ext.∠ABC)=90°

∠PBC+∠QBC=90°

(PB bisect Interior ∠B, QB bisects ext.∠B)

∠PBQ=90°

Similarly, ∠PCQ=90°

Sum of angles of quadrilateral PBCQ =360°

∠BPC+∠PBQ+∠PCQ+∠BQC=360°

∠BPC+∠BQC=180°

∴∠BPQ+∠BQC = 2 rt. angles

✍ʜᴏᴘᴇ ɪᴛ's ʜᴇʟᴘғᴜʟ ᴛᴏ ʏᴏᴜ ✍

Answer 2

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

Given that

BP bisects ∠ABC.

⟹ ∠ABP = ∠CBP = x say

Also,

CP bisects ∠ACB

⟹ ∠ACP = ∠BCP = y say

Also,

BQ bisects ∠DBC

⟹ ∠DBQ = ∠CBQ = z say

Also,

CQ bisects ∠BCE

⟹ ∠ECQ = ∠BCQ = w say

Now,

ABD is a line

⟹ ∠ABC + ∠CBD = 180°

⟹ 2x + 2z = 180°

⟹ x + z = 90°

⟹ ∠PBQ = 90° ------(1)

Again,

ACE is a line

⟹ ∠ACB + ∠BCE = 180°

⟹ 2y + 2w = 180°

⟹ y + w = 90°

⟹ ∠QCP = 90° --------(2)

Now,

In quadrilateral BPCQ

⟹ ∠BPC + ∠PBQ + ∠BQC + ∠QCP = 360°

⟹ ∠BPC + 90° + ∠BQC + 90° = 360°

$\: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \:using \: (1) \: and \: (2) \bigg \}}$

⟹ ∠BPC + ∠BQC + 180° = 360°

⟹ ∠BPC + ∠BQC = 360° - 180°

⟹ ∠BPC + ∠BQC = 180°

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.