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.)
APPEARS IN
संबंधित प्रश्न
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Complete the hexagonal and star shaped rangolies (see the given figures) by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?

In the following figure ; AC = CD; ∠BAD = 110o and ∠ACB = 74o.
Prove that: BC > CD.
In the following figure, write BC, AC, and CD in ascending order of their lengths.
Arrange the sides of the following triangles in an ascending order:
ΔDEF, ∠D = 38°, ∠E = 58°.
ΔABC is isosceles with AB = AC. If BC is extended to D, then prove that AD > AB.
ABCD is a trapezium. Prove that:
CD + DA + AB > BC.
In ΔABC, BC produced to D, such that, AC = CD; ∠BAD = 125° and ∠ACD = 105°. Show that BC > CD.
In ΔPQR, PS ⊥ QR ; prove that: PQ + PR > QR and PQ + QR >2PS.
In ΔABC, D is a point in the interior of the triangle. Prove that DB + DC < AB + AC.
