Question

Prove that the angle included between the bisectors of the vertical angle of a triangle and the perpendicular drawn from the vertex to the base is equal to half the difference of the angles at the base.

Answer 1

Answer:

Answer :

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

To prove: 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

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,

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

…..(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.

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.