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Topic: As good as it gets
Replies: 1   Last Post: Dec 10, 2010 1:24 PM

 Messages: [ Previous | Next ]
 Sujeet Kumar Posts: 190 From: Patna Registered: 10/21/10
Re: As good as it gets
Posted: Dec 10, 2010 1:24 PM

> Only condition to feel the magic of following problem
> is to stick to pure geometry-
>
> Q.In a triangle ABC, point P is taken so that angle
> PAB=30 degree, angle PAC=80 degree, angle PCA=20
> degree and angle PCB=20 degree.Find angle PBC.

I am giving its solution-
Q.In a triangle ABC, point P is taken so that angle PAB=30 degree, angle PAC=80 degree, angle PCA=20 degree and angle PCB=20 degree.Find angle PBC.
Sol.-Contruct an equilateral triangle BMC, so that M is on the same side of BC as of A.Join AM.
Considering triangles BAM and BAC.
BA is common side.
Angle ABM=angle ABC=30 degree.
BC=BM
So, triangle BAM is congruent to BAC.
So, angle BMA=40 degree.
Angle AMC=60-40 =20 degree=angle ACM
AM=AC
Angle CAP=angle CPA=80 degree.
Hence, AC=PC
Considering triangles, PCB and ACM
PC=AC
Angle PCB=angle ACM=20 degree.
BC=MC.
Hence, triangle PCB is congruent to triangle ACM.
So, angle PBC=20 degree.

I gave this problem different title because I was most thrilled after solving this problem, not because
I use different concept for solving this problem but because final diagram that came out looked really
beautiful because of symmetry of the triangles.And I was reminded of the following problem, which
again is very popular one.This is one of the 5 Hardy's perennial along with Langley problem as given
in the book ?Geometry Revisited? by H.S.M. Coxeter and S.L. Greitzer.

Q.In a square ABCD point M is taken so that MA=MB and angle MAB=15 degree.Find angle CMD.

Date Subject Author
12/1/10 Sujeet Kumar
12/10/10 Sujeet Kumar