1) Prove that sum of any two sides of a triangle is greater than twice the median with respect to third side.
2) Show that the sum of three altitudes of a triangle is less than the sum of all three sides of triangle.
3) Prove that any two sides of a triangle are together greater than twice the median drawn to the 3rd side.
4) Prove that the perimeter of a triangle is greater than the sum of its medians.
5) In a quadrilateral ABCD in which diagonals AC BD intersects at O. Show that AB+BC+CD+DA 2(AC+BD)
6) ABC is a triangle in which angle B = twice angle C. D is a point on side BC such that AD bisects angle A and AB= CD. Prove that Angle BAC=72
1) let the median be AD
extend the median to a point E such that AD = ED
you can prove triangle triangle ABD and triangle DEC equal using SAS congruenct
AB = EC (c.p.c.t.)
AC + CE is greater tha AE (sum of two sides of a triangle)
this implies that AC + AB is greater than AE
therefore, AB +AC is greater than AE
2) in a triangle pqr, let the altitudes LMN
we know that hypotenuse is the largest side of a right angled triangle
so, PR is greater than PL .........1
QR is greater than RN ..........2
PQ is greater than QM ..........3
adding 1,2,3 we get
PR + QR + PQ is greater than pl + RN + QN
3) same as the first one
5) i think the question should be 2(AB + BC + CD + DA) is greater than 2(AC + BD)
6)this question is in R.D. sharma and the solutions are provided after the exercise