Question

In a triangle ABC, if AD is the median, then show that ${\text{A}}{{\text{B}}^2}{\text{ + A}}{{\text{C}}^2}{\text{ = 2 (A}}{{\text{D}}^2}{\text{ + B}}{{\text{D}}^2})$

Answer 1

Hint: In this question we will do some construction and also apply the Pythagoras theorem to solve the given problem.

Complete step-by-step answer:
Now, to solve this problem we will construct a perpendicular from vertex A to base BC. Drawing AE perpendicular to BC.

Now we will apply the Pythagorean theorem in right – angled triangles ABE and AEC.
Pythagorean theorem states that in a right – angled triangle the sum of square of base and perpendicular is equal to the square of hypotenuse on the largest side of the triangle. So, in a triangle ABC, right – angled at C,
${\text{A}}{{\text{B}}^2}{\text{ = A}}{{\text{C}}^2}{\text{ + B}}{{\text{C}}^2}$
Where AC is the perpendicular and BC is the base of the triangle ABC.
Now, applying Pythagoras theorem in triangle ABE, we get
${\text{A}}{{\text{B}}^2}{\text{ = B}}{{\text{E}}^2}{\text{ + A}}{{\text{E}}^2}$ …… (1)
Applying Pythagoras theorem in triangle AEC,
${\text{A}}{{\text{C}}^2}{\text{ = A}}{{\text{E}}^2}{\text{ + E}}{{\text{C}}^2}$ ……. (2)
Now, adding equation (1) and (2),
${\text{A}}{{\text{B}}^2}{\text{ + A}}{{\text{C}}^2}{\text{ = A}}{{\text{E}}^2}{\text{ + B}}{{\text{E}}^2}{\text{ + E}}{{\text{C}}^2}{\text{ + A}}{{\text{E}}^2}$
${\text{A}}{{\text{B}}^2}{\text{ + A}}{{\text{C}}^2}{\text{ = 2A}}{{\text{E}}^2}{\text{ + B}}{{\text{E}}^2}{\text{ + E}}{{\text{C}}^2}$
BE = BD – ED and EC = ED + DC
${\text{A}}{{\text{B}}^2}{\text{ + A}}{{\text{C}}^2}{\text{ = 2A}}{{\text{E}}^2}{\text{ + (BD - ED}}{{\text{)}}^2}{\text{ + (ED + DC}}{{\text{)}}^2}$
Solving the above equation, we get
${\text{A}}{{\text{B}}^2}{\text{ + A}}{{\text{C}}^2}{\text{ = 2A}}{{\text{E}}^2}{\text{ + B}}{{\text{D}}^2}{\text{ + E}}{{\text{D}}^2}{\text{ - 2BDED + E}}{{\text{D}}^2}{\text{ + D}}{{\text{C}}^2}{\text{ + 2EDDC}}$
Now, AD is the median so, BD = DC
So,
${\text{A}}{{\text{B}}^2}{\text{ + A}}{{\text{C}}^2}{\text{ = 2A}}{{\text{E}}^2}{\text{ + B}}{{\text{D}}^2}{\text{ + E}}{{\text{D}}^2}{\text{ - 2DCED + E}}{{\text{D}}^2}{\text{ + D}}{{\text{C}}^2}{\text{ + 2EDDC}}$
${\text{A}}{{\text{B}}^2}{\text{ + A}}{{\text{C}}^2}{\text{ = 2A}}{{\text{E}}^2}{\text{ + B}}{{\text{D}}^2}{\text{ + E}}{{\text{D}}^2}{\text{ + E}}{{\text{D}}^2}{\text{ + D}}{{\text{C}}^2}$
${\text{A}}{{\text{B}}^2}{\text{ + A}}{{\text{C}}^2}{\text{ = 2(A}}{{\text{E}}^2}{\text{ + B}}{{\text{D}}^2}{\text{ + E}}{{\text{D}}^2})$ as BD = DC …. (3)
Now from triangle AED,
${\text{A}}{{\text{D}}^2}{\text{ = A}}{{\text{E}}^2}{\text{ + E}}{{\text{D}}^2}$
Putting this value in equation (3), we get
${\text{A}}{{\text{B}}^2}{\text{ + A}}{{\text{C}}^2}{\text{ = 2(A}}{{\text{D}}^2}{\text{ + B}}{{\text{D}}^2})$
Hence proved.

Note: While solving such types of questions, you have to think how you can solve the given problem. By carefully seeing the figure given and understanding the question you can easily know which property you have to use to solve the question. Also, carefully apply the property you use to get the correct answer.