Advertisements
Advertisements
प्रश्न
D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC.
Advertisements
उत्तर
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
संबंधित प्रश्न
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
In the following figure, ∠BAC = 60o and ∠ABC = 65o.
Prove that:
(i) CF > AF
(ii) DC > DF
In the following figure, write BC, AC, and CD in ascending order of their lengths.
“Caste inequalities are still prevalent in India.” Examine the statement.
Name the greatest and the smallest sides in the following triangles:
ΔABC, ∠ = 56°, ∠B = 64° and ∠C = 60°.
Name the greatest and the smallest sides in the following triangles:
ΔDEF, ∠D = 32°, ∠E = 56° and ∠F = 92°.
In ΔABC, BC produced to D, such that, AC = CD; ∠BAD = 125° and ∠ACD = 105°. Show that BC > CD.
In the given figure, T is a point on the side PR of an equilateral triangle PQR. Show that PT < QT
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.
