Kindly answer the question as provided below and explain each and every part of the solution.......
Solution:
We are given a triangle ABC and AD is median on the side BC.
We can prove AB + BC + CA > 2AD by starting with the triangles.
So, in the triangle ABC. We have sub-triangles ABD and ADC.
So, in triangle ABD
Using the inequality of the triangle, the sum of any two sides is always greater than or equal to the third side.
We have AB + BD > AD .....(1)
Also using the same in triangle ADC,
We have AC + DC > AD .....(2)
Adding equations (1) and (2), We get
AB + AC + (BD + DC) > 2AD
⇒ AB + AC + BC > 2AD
Hence proved.
We are given a triangle ABC and AD is median on the side BC.
We can prove AB + BC + CA > 2AD by starting with the triangles.
So, in the triangle ABC. We have sub-triangles ABD and ADC.
So, in triangle ABD
Using the inequality of the triangle, the sum of any two sides is always greater than or equal to the third side.
We have AB + BD > AD .....(1)
Also using the same in triangle ADC,
We have AC + DC > AD .....(2)
Adding equations (1) and (2), We get
AB + AC + (BD + DC) > 2AD
⇒ AB + AC + BC > 2AD
Hence proved.