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
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 a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
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?

From the following figure, prove that: AB > CD.

ΔABC is isosceles with AB = AC. If BC is extended to D, then prove that AD > AB.
Prove that the hypotenuse is the longest side in a right-angled triangle.
D is a point on the side of the BC of ΔABC. Prove that the perimeter of ΔABC is greater than twice of AD.
In ΔPQR is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.
Prove that in an isosceles triangle any of its equal sides is greater than the straight line joining the vertex to any point on the base of the triangle.
