**The Area Way**

Incases , where a relation with areas is possible, It is quite efficient to use it to solve the problem.

Example –

If are the altitudes of a triangle from A,B,C to the opposite sides and r is the inradius. Prove that –

1.

2.

**Proof**

Let I be the incentre.

Note that ,

and that ,

Hence proved.

2. By Cauchy ,

And using the identity, we are done.

3. Let ABC be a triangle with H as its orthocentre , and AD,BE,CF its altitudes.

Prove that –

**Proof**

Both AD and HD are altitudes to the same base BC , Hence the areas of triangles ABC and HBC are in the ratio of their altitudes (*)
Let,

Thus , the inequality is equivalent to –

Note that –

Hence it is true by Cauchy-Schwarz Inequality.

4. With the Same notations prove that –

**Proof**

Note that –

$latex \frac{HD}{HA} = \frac{HD}{AD – HD} = \frac{S_1}{S-S_1} = \frac{S_1}{S_2+S_3}$

Thus ,

The last inequality is the famous Nesbitt’s Inequality. Hence Proved.

Also note that – equality occurs when

Or, when the triangle is equilateral.

**Exercises **

1. ABC triangle, L,M,N are points on BC,CA,AB and P,Q,R are intersections of AL,BM,CN with the circum-circle of the triangle. Prove that –

*(Source : Korea 1995)*

2. Let AD,BE,CF be altitudes from A,B,C of triangle ABC and PQ , PR, PS be projections of P on BC,CA,AB . Prove that –

3. The altitudes AD , BE , CF intersect the circum-circles at D’ , E’ , F’ . The prove that –

(a)

(b)

**The Equilateral Usage**

In almost every geometrical inequality, we would note that , the equality condition refers to the equilateral condition – triangle or any polygon.

There has to be some way of utilising this fact right ? , Well there IS .

We can illustrate by a different and beautiful proof of our Weitzenbock’s Inequality –

For a triangle ABC , Prove that –

where, is the Area of the triangle.

### Proof

Consider the equilateral triangle of length , It has altitude – and area –

Consider the Given triangle – ABC , with altitude – AD – h.

Let AD divide BC into – and

For our triangle ABC , let , .

Therefore,

with equality occuring only when –