Advertisements
Advertisements
प्रश्न
In the following figure, AD is the bisector of ∠BAC. Prove that AB > BD.
Advertisements
उत्तर
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
संबंधित प्रश्न
In Fig. 10.25, AB = AC and DB = DC, find the ratio ∠ABD : ∠ACD.
Compute the value of x in the following figure:
Is the following statement true and false :
All the angles of a triangle can be equal to 60°.
Is the following statement true and false :
A triangle can have at most one obtuse angles.
Fill in the blank to make the following statement true:
An exterior angle of a triangle is always ......... than either of the interior opposite angles.
State, if the triangle is possible with the following angles :
125°, 40°, and 15°
In the following, find the marked unknown angle:
D is any point on side AC of a ∆ABC with AB = AC. Show that CD < BD.
P is a point on the bisector of ∠ABC. If the line through P, parallel to BA meet BC at Q, prove that BPQ is an isosceles triangle.
How is a vertex of triangle △ABC most accurately defined?
