Question

don't answer if you don't know​

Answer 1

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

In ∆ ABC,

BD and CD are internal angle bisector of ∠ B and ∠C respectively.

∠BAC = y and ∠BDC = x

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:180 \degree \: + \: y \: = \: 2x$

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

Given that,

In ∆ ABC

BD bisects ∠ABC

So, ∠ABD = ∠DBC = 'a' say

Also, CD bisects ∠ACB

So, ∠ACD = ∠BCD = 'b' say.

Now,

In ∆ BCD

We know,

Sum of all interior angles of a triangle is supplementary.

So, ∠DBC + ∠DCB + ∠BCD = 180°

$\rm :\longmapsto\:a + b + x = 180 \degree$

$\bf\implies \:a + b = 180\degree - x - - - (1)$

Now,

In ∆ABC

We know,

Sum of all interior angles of a triangle is supplementary.

So, ∠ABC + ∠ACB + ∠BAC = 180°

$\red{\rm :\longmapsto\:2a + 2b + y = 180\degree}$

$\rm :\longmapsto\:2(a + b) + y = 180\degree$

On substituting the value of a + b, from equation (1), we get

$\rm :\longmapsto\:2(180\degree - x) + y = 180\degree$

$\rm :\longmapsto\:360\degree - 2x + y = 180\degree$

$\rm :\longmapsto\:360\degree + y - 180\degree = 2x$

$\rm :\longmapsto\:180\degree + y = 2x$

Hence, Proved

Additional Information :-

1. Sum of two sides of a triangle is always greater than third side.

2. Angle opposite to longest side is always greater.

3. Side opposite to largest angle is always longest.