In a triangle ABC AD is the bisector of angle BAC. Show that AB>BD and AC>CD

from the figure,*L***BDA**=90^{0},BECAUSE,ADis the bisector.

consider triangle ABD,it's right angled so,BD is the base and AB is the hypotnuse,

we know that,hypt is always greater than base ,so AB**>**BD and similarly,CD is base and AC is hypt,

so AC**>**CD.

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considering an equilateral triangle ABC

with sides equal sides AB ; AC and BC

since AD is the bisector of angle A and D is located between B & C

therefore BC = CD + DB

and CD = DB

BC = 2DB

considering the pythagorean theorem that states;

In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).

given by the equation; a^2 + b^2 = c^2

considering AB as the hypothenuse, AD is the altitude or one of the leg of the triangle, DB one of the leg.

Base from the pythagorean theorem alone AB is the side opposite of the right angle and therefore is the hypothenuse

AB^2 = AD^2 + DB^2

AB = sqr root (AD^2 + DB^2)

therefore AB is greater than DB since the hypothenuse of the triangle is the longest side

with sides equal sides AB ; AC and BC

since AD is the bisector of angle A and D is located between B & C

therefore BC = CD + DB

and CD = DB

BC = 2DB

considering the pythagorean theorem that states;

In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).

given by the equation; a^2 + b^2 = c^2

considering AB as the hypothenuse, AD is the altitude or one of the leg of the triangle, DB one of the leg.

Base from the pythagorean theorem alone AB is the side opposite of the right angle and therefore is the hypothenuse

AB^2 = AD^2 + DB^2

AB = sqr root (AD^2 + DB^2)

therefore AB is greater than DB since the hypothenuse of the triangle is the longest side

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If AD is the bisector of ∠A of triangle ABC, show that AB>DB.

Given AD is the bisector of ∠A of triangle ABC

Hence ∠DAB=∠DAC

∠BDA is the exterior angle of the ∆DAC

Hence ∠BDA> ∠DAC

or ∠BDA > ∠DAB Since ∠DAC=∠DAB

→ AB > BD In a triangle sides opposite to greater angle is greater.

this is the right answer ignore the first

Given AD is the bisector of ∠A of triangle ABC

Hence ∠DAB=∠DAC

∠BDA is the exterior angle of the ∆DAC

Hence ∠BDA> ∠DAC

or ∠BDA > ∠DAB Since ∠DAC=∠DAB

→ AB > BD In a triangle sides opposite to greater angle is greater.

this is the right answer ignore the first

- 85

If AD is the bisector of ∠A of triangle ABC, show that AB>DB.

Given :AD is the bisector of ∠A of triangle ABC .

Hence, ∠DAB=∠DAC

∠BDA is the exterior angle of the ∆DAC.

Hence, ∠BDA> ∠DAC

or ∠BDA > ∠DAB

Since ,∠DAC=∠DAB

→ AB > BD In a triangle sides opposite to greater angle is greater.

Given :AD is the bisector of ∠A of triangle ABC .

Hence, ∠DAB=∠DAC

∠BDA is the exterior angle of the ∆DAC.

Hence, ∠BDA> ∠DAC

or ∠BDA > ∠DAB

Since ,∠DAC=∠DAB

→ AB > BD In a triangle sides opposite to greater angle is greater.

- 1

If AD is the bisector of ∠A of triangle ABC, show that AB>DB.

Given :AD is the bisector of ∠A of triangle ABC .

Hence, ∠DAB=∠DAC

∠BDA is the exterior angle of the ∆DAC.

Hence, ∠BDA> ∠DAC

or ∠BDA > ∠DAB

Since ,∠DAC=∠DAB

→ AB > BD In a triangle sides opposite to greater angle is greater.

Given :AD is the bisector of ∠A of triangle ABC .

Hence, ∠DAB=∠DAC

∠BDA is the exterior angle of the ∆DAC.

Hence, ∠BDA> ∠DAC

or ∠BDA > ∠DAB

Since ,∠DAC=∠DAB

→ AB > BD In a triangle sides opposite to greater angle is greater.

- 1

If AD is the bisector of ?A of triangle ABC, show that AB>DB.?

Given AD is the bisector of ?A of triangle ABC?

Hence ?DAB=?DAC?

?BDA is the exterior angle of the ?DAC?

Hence ?BDA> ?DAC?

or ?BDA > ?DAB Since ?DAC=?DAB?

? AB > BD In a triangle sides opposite to greater angle is greater.

this is the right answer ignore the first

?

Given AD is the bisector of ?A of triangle ABC?

Hence ?DAB=?DAC?

?BDA is the exterior angle of the ?DAC?

Hence ?BDA> ?DAC?

or ?BDA > ?DAB Since ?DAC=?DAB?

? AB > BD In a triangle sides opposite to greater angle is greater.

this is the right answer ignore the first

?

- 1

If AD is the bisector of ∠A of triangle ABC, show that AB>DB.

Given :AD is the bisector of ∠A of triangle ABC .

Hence, ∠DAB=∠DAC

∠BDA is the exterior angle of the ∆DAC.

Hence, ∠BDA> ∠DAC

or ∠BDA > ∠DAB

Since ,∠DAC=∠DAB

→ AB > BD In a triangle sides opposite to greater angle is greater.

Given :AD is the bisector of ∠A of triangle ABC .

Hence, ∠DAB=∠DAC

∠BDA is the exterior angle of the ∆DAC.

Hence, ∠BDA> ∠DAC

or ∠BDA > ∠DAB

Since ,∠DAC=∠DAB

→ AB > BD In a triangle sides opposite to greater angle is greater.

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