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 the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that ∠A > ∠C and ∠B > ∠D.
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.
Arrange the sides of the following triangles in an ascending order:
ΔABC, ∠A = 45°, ∠B = 65°.
Name the smallest angle in each of these triangles:
In ΔABC, AB = 6.2cm, BC = 5.6cm and AC = 4.2cm
Name the smallest angle in each of these triangles:
In ΔPQR, PQ = 8.3cm, QR = 5.4cm and PR = 7.2cm
Prove that the perimeter of a triangle is greater than the sum of its three medians.
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PR > PS
In ΔPQR is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.
