Advertisements
Advertisements
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.
Advertisements
उत्तर
In the given figure,
∠ABC + ∠PBC = 180° (Linear pair)
⇒ ∠ABC = 180° − ∠PBC ... (1)
Also,
∠ACB + ∠QCB = 180°
∠ACB = 180° − ∠QCB … (2)
As ∠PBC < ∠QCB,
⇒ 180º − ∠PBC > 180º − ∠QCB
⇒ ∠ABC > ∠ACB [From equations (1) and (2)]
⇒ AC > AB (Side opposite to the larger angle is larger.)
संबंधित प्रश्न
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
From the following figure, prove that: AB > CD.
In a triangle PQR; QR = PR and ∠P = 36o. Which is the largest side of the triangle?
If two sides of a triangle are 8 cm and 13 cm, then the length of the third side is between a cm and b cm. Find the values of a and b such that a is less than b.
In the following figure, write BC, AC, and CD in ascending order of their lengths.
D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC.
D is a point on the side of the BC of ΔABC. Prove that the perimeter of ΔABC is greater than twice of AD.
In ABC, P, Q and R are points on AB, BC and AC respectively. Prove that AB + BC + AC > PQ + QR + PR.
In ΔPQR, PS ⊥ QR ; prove that: PQ + PR > QR and PQ + QR >2PS.
In the given figure, T is a point on the side PR of an equilateral triangle PQR. Show that RT < QT
