Advertisements
Advertisements
Question
In the following figure, AD is the bisector of ∠BAC. Prove that AB > BD.

Advertisements
Solution
Given: ABC is a triangle such that AD is the bisector of ∠BAC.
To prove: AB > BD.
Proof: Since, AD is the bisector of ∠BAC.
But ∠BAD = CAD ...(i)
∴ ∠ADB > ∠CAD ...[Exterior angle of a triangle is greater than each of the opposite interior angle]
∴ ∠ADB > ∠BAD ...[From equation (i)]
AB > BD ...[Side opposite to greater angle is longer]
Hence proved.
APPEARS IN
RELATED QUESTIONS
Is the following statement true and false :
All the angles of a triangle can be less than 60°
Define a triangle.
The bisects of exterior angle at B and C of ΔABC meet at O. If ∠A = x°, then ∠BOC =
Calculate the angles of a triangle if they are in the ratio 4: 5: 6.
One angle of a triangle is 61° and the other two angles are in the ratio `1 1/2: 1 1/3`. Find these angles.
Can a triangle together have the following angles?
33°, 74° and 73°
One angle of a right-angled triangle is 70°. Find the other acute angle.
In a ∆ABC, AB = AC. The value of x is ________
What is common in the following figure?
![]() |
![]() |
| (i) | (ii) |
Is figure (i) that of triangle? if not, why?
How is a vertex of triangle △ABC most accurately defined?


