Geometry- a powerful technique

Original Post.

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.

Let ABCD be a quadrilateral whose AC & BD meet at O, such that ABO and BCO are similar. Then if AO = a, BO = b; we can write CO and DO in terms of a,b,k where k 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, …..

Here I am representing a set of problems which I solved using the method:
1. (RMO-1992; Theorem 1 in my opinion :P ) Let ABCD be a cyclic quadrilateral such that AC\perp BD; AC meets BD at E. Prove that
EA^2 + EB^2 + EC^2 + ED^2 = 4R^2
where R is the radius of the circumscribing circle.

As ABCD is cyclic, \triangle AED \sim \triangle BEC\implies \frac {EC}{EB} = \frac {ED}{EA} = k(say).
Letting EA = a, EB = b\implies EC = kb, ED = ka
Now, AD^2 = k^2a^2 + a^2 = a^2(k^2 + 1) and AB^2 = a^2 + b^2.
Now from \triangle ADB, Sine rule gives:
2R = \frac {AB}{\sin\angle ADB}\implies 4R^2 = \frac {AB^2}{\sin\angle ADE}
\therefore 4R^2 = \frac {a^2 + b^2}{\frac {a^2}{(1 + k^2)a^2}} = (1 + k^2)(a^2 + b^2)
But EA^2 + EB^2 + EC^2 + ED^2 = a^2 + b^2 + k^2b^2 + k^2a^2
= (1 + k^2)(a^2 + b^2)
Comparing these yields the result.
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 K,L \to respective midpoints of AB, CD respectively. Prove that OKML is a parallelogram.
(Sorry for my delay in updating)

Let AM = a, BM = b
Now \triangle AMB\sim \triangle CMD as ABCD is a cyclic quadrilateral.
Hence \frac {DM}{AM} = \frac {CM}{BM} = k (say)
So that DM = ak; CM = bk
AS ABM is a right triangle, KB = \frac {1}{2}AB = \frac {1}{2}\sqrt {a^2 + b^2}
AS CDM is a right triangle, CL = \frac {1}{2}CD = \frac {k}{2}\sqrt {a^2 + b^2}
Hence in \triangle BKM and \triangle CML;
\frac {BM}{MC} = \frac {BK}{LC} = \frac {1}{k};
\angle ACD = \angle ABD\implies \angle MCL = \angle MBK
So that \triangle MBK \sim \triangle CLM
\therefore \angle MKB - 90^{\circ} = \angle MLC - 90^{\circ}
\therefore\boxed{\angle MKO = \angle MLO}
Now; triangle AMK gives:
\cos \angle MAK = \frac {MK^2 - a^2 - \frac {1}{4}(a^2 + b^2)}{2a\frac {1}{2}\sqrt {a^2 + b^2}}
So MK^2 = a^2 + \frac {1}{4}(a^2 + b^2) - a\sqrt {a^2 + b^2}\cos \angle MAK
= a^2 + \frac {1}{4}(a^2 + b^2) - a\sqrt {a^2 + b^2}\cos \angle MDC
= a^2 + \frac {1}{4}(a^2 + b^2) - a\sqrt {a^2 + b^2}\cdot \frac {ak}{k\sqrt {a^2 + b^2}}
= \frac {1}{4}(a^2 + b^2)
So \boxed{MK = \frac {1}{2}\sqrt {a^2 + b^2}}
From the problem before, we know that R^2 = \frac {(k^2 + 1)(a^2 + b^2)}{4}
CL = \frac {k}{2}\sqrt {a^2 + b^2}; OC^2 = R^2 = \frac {(k^2 + 1)(a^2 + b^2)}{4}
So OL^2 = \frac {1}{4}(a^2 + b^2)\implies \boxed{OL = \frac {1}{2}\sqrt {a^2 + b^2}}
So we have; \boxed{\mathbf{MK = OL}}

Hence from OKML; .
ML = OL; \angle MKO = \angle MLO
So that OKML is a ||^{\text{gm}}
3.(RMO-1997, India) In quadr. ABCD; AB||CD & AC\perp BD. Show that :
(a)AD\cdot BC\ge AB\cdot CD
(b)AD + BC\ge AB + CD
Solution:(by OCHA)

Let ABCD be the quadr.
\triangle AOD\sim \triangle DOC
So that \frac{OC}{AO}=\frac{OD}{BO}=k (say)
Letting AO=a, BO=b; \implies OC=ka, OD=kb
We have some relations:
AD^2\cdot BC^2=a^2b^2+k^2a^4+k^2b^2+k^4a^2b^2
=a^2b^2(1+k^4)+(ka^2)^2+(kb^2)^2\ge 2a^2b^2k^2+(ka^2)^2+(kb^2)^2
=k^2(a^2+b^2)^2=AB^2\cdot CD^2
\therefore \boxed{AD\cdot BC\ge AB\cdot CD}
Also; AD^2+BC^2=AB^2+CD^2=(a^2+b^2)(1+k^2)
\implies AD^2+BC^2+2AD\cdot BC\geq AB^2+CD^2+2AB\cdot CD
\implies (AD+BC)^2\ge (AB+CD)^2
\boxed{\implies AD+BC\geq AB+CD}


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