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 a ΔABC, if ∠A = 55°, ∠B = 40°, find ∠C.
Compute the value of x in the following figure:
Fill in the blank to make the following statement true:
Sum of the angles of a triangle is ....
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.
In the given figure, side BC of ΔABC is produced to point D such that bisectors of ∠ABC and ∠ACD meet at a point E. If ∠BAC = 68°, find ∠BEC.
Side BC of a triangle ABC has been produced to a point D such that ∠ACD = 120°. If ∠B = \[\frac{1}{2}\]∠A is equal to
In a triangle PQR, ∠P = 60° and ∠Q = ∠R, find ∠R.
In the given figure, express a in terms of b.
If an angle of a triangle is equal to the sum of the other two angles, find the type of the triangle
In the following figure, PQ ⊥ RQ, PQ = 5 cm and QR = 5 cm. Then ∆PQR is ______.
