Advertisements
Advertisements
प्रश्न
In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.
Advertisements
उत्तर
As PR > PQ,
∴ ∠PQR > ∠PRQ (Angle opposite to larger side is larger) ... (1)
PS is the bisector of ∠QPR.
∴∠QPS = ∠RPS ... (2)
∠PSR is the exterior angle of ΔPQS.
∴ ∠PSR = ∠PQR + ∠QPS ... (3)
∠PSQ is the exterior angle of ΔPRS.
∴ ∠PSQ = ∠PRQ + ∠RPS ... (4)
Adding equations (1) and (2), we obtain
∠PQR + ∠QPS > ∠PRQ + ∠RPS
⇒ ∠PSR > ∠PSQ [Using the values of equations (3) and (4)]
APPEARS IN
संबंधित प्रश्न
In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that ∠A > ∠C and ∠B > ∠D.
In a triangle PQR; QR = PR and ∠P = 36o. Which is the largest side of the triangle?
In the following figure, write BC, AC, and CD in ascending order of their lengths.
Arrange the sides of ∆BOC in descending order of their lengths. BO and CO are bisectors of angles ABC and ACB respectively.
In ΔABC, the exterior ∠PBC > exterior ∠QCB. Prove that AB > AC.
Prove that the hypotenuse is the longest side in a right-angled triangle.
In ABC, P, Q and R are points on AB, BC and AC respectively. Prove that AB + BC + AC > PQ + QR + PR.
In ΔPQR, PR > PQ and T is a point on PR such that PT = PQ. Prove that QR > TR.
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that : PN < RN
