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

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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but according to the figure given ABD is not a right angled triangle. And also give me the difference between bisector and perpendicular bisector

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thais don't u know the difference b/w perpendicular bisector and bisector???

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AD is the bisector angleBAC prove that AB is greater than BD and AC is greater than 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 = 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

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????
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I agree with abhinash john
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AD is bisector of angle BAC Therefore angle BAD= angle CAD Let angle BAD= angle CAD =x In triangle ADC by exterior angle property X+angle C = angle ADB Therefore angle ADB is greater than x By side opp. To equal angle are equal So, AB is greater than BD
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This is the answer but for proving AB>BD similarly we can also prove AC>CD • 19 Hope it helps u!!!
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therefore <DAB=<DAC
To Prove:-AB>BD
But,<DAB=<DAC
Therefore,AB >BD(Opposite side of the greatest angle is greater)

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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.

• 1
????
• -4
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.

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Gjdbbe fb
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yes hindu ish is correct

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jhdg
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golu is correct
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what is this
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The answers are right but do send a diagram with your question if you are dealing with geometry for better understanding.
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sorry
• 0
This is question of inequalities
So lr formula(sum of two angles is always greater than one angle)
And solve it
• 1 • 0
How can we prove ac>cd? Can someone explain that also
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Worked really hard to find the solution. Hope this helps you all. All the best and move on.👍 • 2
Heyy! The answer is as follows: • 0
Send question one more time
• 0 • 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
?
• 1 • 1 • 0 • 0

AB>BD(side opposite to greater angle is longer)
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Congruent part of congruent triangle
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ttte5etteye6dyyrj43rd
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The criteria that is used is:Sides opposite to the larger angles are longer. • 0 • 0
Bisector of angle divides angle into two equal parts . And in perpendicular bisector the line bisects a angle as well as when it falls on a line it makes 90 ?
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Its easy
• 0 • 0 • 0 • 0
use angle sum property
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Considering AB as the hypothenuse, AD is the altitude or one of the leg of the triangle, DB one of the leg. If AD is the bisector of ∠A of triangle ABCshow that AB>DB. → AB > BD In a triangle sides opposite to greater angle is greater
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Wryu
• 0 • -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.

• 0
CA BD
• 0
bhai Kismat acchi hai ki acchi hai
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