In AABC, D is any point on the side BC. Show that AB + BC + CA>2AD
Answers
Given :-
- In triangle ABC, D is any point on BC.
To Show :-
- AB + BC + CA > 2 AD
Proof :-
In triangle ABD,
- We know that, sum of two sides of a triangle is greater than third side.
So,
- AB + BD > AD -----(1)
In triangle ACD
- Sum of two sides of a triangle is greater than third side.
So,
- AC + CD > AD -----(2)
☆ On adding equation (1) and equation (2), we get
AB + BD + AC + CD > AD + AD
AB + (BD + CD) + AC > 2 AD
AB + BC + AC > 2 AD
Hence, Proved.
Additional Information :-
Properties of a triangle
A triangle has three sides, three angles, and three vertices.
The sum of all internal angles of a triangle is always equal to 180°. This is called the angle sum property of a triangle.
The sum of the length of any two sides of a triangle is greater than the length of the third side.
The side opposite to the largest angle of a triangle is the largest side.
Any exterior angle of the triangle is equal to the sum of its interior opposite angles. This is called the exterior angle property of a triangle.
Based on the angle measurement, there are three types of triangles:
Acute Angled Triangle : A triangle in which all three angles are less than 90° is an acute angle triangle.
Right-Angled Triangle : A triangle in which one angle is 90° is a right-angle triangle.
Obtuse Angled Triangle : A triangle that has one angle more than 90° is an obtuse angle triangle.
