Advertisements
Advertisements
प्रश्न
From the following figure, prove that: AB > CD.
Advertisements
उत्तर
In ΔABC,
AB = AC ...[ Given ]
∴ ∠ACB = ∠B ...[ angles opposite to equal sides are equal ]
∠B = 70° ...[ Given ]
⇒ ∠ACB = 70° ...(i)
Now,
∠ACB +∠ACD = 180°...[ BCD is a straight line]
⇒ 70° + ∠ACD = 180°
⇒ ∠ACD = 110° ...(ii)
In ΔACD,
∠CAD + ∠ACD + ∠D = 180°
⇒ ∠CAD + 110° + ∠D = 180° ...[ From (ii) ]
⇒ ∠CAD + ∠D = 70°
But ∠D = 40° ...[ Given ]
⇒ ∠CAD + 40°= 70°
⇒ ∠CAD = 30° …(iii)
In ΔACD,
∠ACD = 110° ...[ From (ii) ]
∠CAD = 30° ...[ From (iii) ]
∠D = 40° ...[ Given ]
∴ D > ∠CAD
⇒ AC > CD ....[Greater angle has greater side opposite to it]
Also,
AB = AC ...[ Given ]
Therefore, AB > CD.
संबंधित प्रश्न
Show that in a right angled triangle, the hypotenuse is the longest side.
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.
In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.
Arrange the sides of ∆BOC in descending order of their lengths. BO and CO are bisectors of angles ABC and ACB respectively.
D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC.
Name the smallest angle in each of these triangles:
In ΔABC, AB = 6.2cm, BC = 5.6cm and AC = 4.2cm
Prove that the perimeter of a triangle is greater than the sum of its three medians.
ABCD is a quadrilateral in which the diagonals AC and BD intersect at O. Prove that AB + BC + CD + AD < 2(AC + BC).
In ΔPQR is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.
In ΔABC, AE is the bisector of ∠BAC. D is a point on AC such that AB = AD. Prove that BE = DE and ∠ABD > ∠C.
