Advertisements
Advertisements
Question
D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC.
Advertisements
Solution
In ΔADC
AD > AC
⇒ ∠ACD > ∠ADC ...(i)
In ΔABD
∠ABD + ∠BAD = ∠ADC ...(ii)
Putting the value of ∠ADC in equation (i)
∠ACD > ∠ABD + ∠BAD ...(iii)
⇒ ∠ACD > ∠ABD
⇒ ∠ACB > ∠ABC
AB > AC
APPEARS IN
RELATED QUESTIONS
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
In the following figure ; AC = CD; ∠BAD = 110o and ∠ACB = 74o.
Prove that: BC > CD.
Name the greatest and the smallest sides in the following triangles:
ΔDEF, ∠D = 32°, ∠E = 56° and ∠F = 92°.
Name the smallest angle in each of these triangles:
In ΔXYZ, XY = 6.2cm, XY = 6.8cm and YZ = 5cm
In ΔABC, the exterior ∠PBC > exterior ∠QCB. Prove that AB > AC.
Prove that the perimeter of a triangle is greater than the sum of its three medians.
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 is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.
In ΔABC, D is a point in the interior of the triangle. Prove that DB + DC < AB + AC.
