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

The answer provided above by sahil sharma is correct . In case you need any extra help or need extra explanation you can get back to us.

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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

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