Question

Let's prove that the angle between the bisector of the vertical
angle of a triangle and the perpendicular drawn from the
vertex to the base of the triangle is half of the difference of
the base angles​

Answer 1

Answer:

Let the triangle be ΔABC, Let AE be the bisector of vertical angle A, And Let AD be the perpendicular drawn from the vertex A to the base BC of the triangle ABC. ... Here ∠ABC and ∠ACB are base angles and ∠DAE is the angle between the bisector of vertical angle and perpendicular from vertex to the base.

Answer 2

Answer:

$\huge\mathcal\blue{Given:}$

a triangle with vertical angle bisector, perpendicular from vertex to the base

$\huge\mathcal\blue{To find:}$

the angle between the bisector of the vertical angle of a triangle and the perpendicular drawn from the vertex to the base of the triangle is half of the difference of the base angles

$\huge\mathcal\blue{Solution:}$

Let the triangle be ΔABC,

Let AE be the bisector of vertical angle A,

And Let AD be the perpendicular drawn from the vertex A to the base BC of the triangle ABC.

So the corresponding figure will be as shown below,

(see attached)

Now in ΔABC,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠BAC + ∠ABC + ∠ACB = 180°………(i)

Given AE is angle bisector of ∠BAC

⇒ ∠BAE = ∠CAE

⇒ ∠BAC = 2∠BAE

Substituting the above value in equation (i) we get

2∠BAE + ∠ABC + ∠ACB = 180°-----(ii)

But from figure, ∠BAE = ∠BAD + ∠DAE

Substituting this value in equation (ii), we get

2(∠BAD + ∠DAE) + ∠ABC + ∠ACB = 180°

⇒ 2∠BAD + 2∠DAE + ∠ABC + ∠ACB = 180°-----(iii)

Given AD is perpendicular to BC, so ΔBAD and ΔDAE is right - angled triangle, hence

In right - angled ΔBAD,

∠ABD + ∠BAD = 90°

⇒ ∠ABC + ∠BAD = 90°⇒ ∠BAD = 90° - ∠ABC-----(iv)

Substituting equation (iv) in equation (iii), we get

⇒ 2(90° - ∠ABC) + 2∠DAE + ∠ABC + ∠ACB = 180°

⇒ 180° - 2∠ABC + 2∠DAE + ∠ABC + ∠ACB = 180°

⇒ 180° - ∠ABC + 2∠DAE + ∠ACB = 180°

⇒ 2∠DAE = 180° - 180° + ∠ABC - ∠ACB

⇒ 2∠DAE = ∠ABC - ∠ACB

$< DAE = \frac{1}{2}( < ABC - < ACB)-----(v)$

Here ∠ABC and ∠ACB are base angles and ∠DAE is the angle between the bisector of vertical angle and perpendicular from vertex to the base.

Hence the angle between the bisector of the vertical angle of a triangle and the perpendicular drawn from the vertex to the base of the triangle is half of the difference of the base angles.

Hence Proved.