I am writing this article from the little experience & knowledge that I have gained from solving the RMO & INMO Geometry problems of our country. However, I have solved more than 7 problems in Geometry using this powerful method, which is completely based on similarity of triangles.

I am sorry that I could not post any topic due to my exams before.

A method related to similarity of triangles

We face several problems related to similarity of triangles. In some cases, similarity is used in proving theorems related to cyclic quadrilaterals. Now I come to my method.

**Statement**

Let ABCD be a quadrilateral whose AC & BD meet at O, such that ABO and BCO are similar. Then if we can write and in terms of where is the scaling factor of the two triangles.

The triangles are similar if:

1) It is a cyclic quadrilateral;

2) If its opposite sides are parallel , ie if it is a trapezoid;

In most problems, we are given that AC and BD are perpendicular, leading to the experessions of AB, CD, …..

**APPLICATIONS**

Here I am representing a set of problems which I solved using the method:

1. (RMO-1992; Theorem 1 in my opinion ) Let ABCD be a cyclic quadrilateral such that ; AC meets BD at E. Prove that

where R is the radius of the circumscribing circle.

*Solution:*

As is cyclic, (say).

Letting

Now, and .

Now from , Sine rule gives:

But

Comparing these yields the result.

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2.Let AC and BD be chords of a circle with centre O such that the intersect at right angles inside the circle at point M. Suppose respective midpoints of respectively. Prove that is a parallelogram.

*Solution:*

(Sorry for my delay in updating)

Let

Now as ABCD is a cyclic quadrilateral.

Hence (say)

So that

AS ABM is a right triangle,

AS CDM is a right triangle,

Hence in and ;

So that

Now; triangle AMK gives:

So

So

From the problem before, we know that

So

So we have;

Hence from ; .

So that is a

QED’

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3.(RMO-1997, India) In quadr. ABCD; & . Show that :

(a)

(b)

Solution:(by OCHA)

Let ABCD be the quadr.

So that (say)

Letting

We have some relations:

Hence

Also;

QED