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Topic: Yet another exciting geometry problem
Replies: 1   Last Post: Dec 19, 2010 10:54 AM

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Sujeet Kumar

Posts: 190
From: Patna
Registered: 10/21/10
Re: Yet another exciting geometry problem
Posted: Dec 19, 2010 10:54 AM
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> Here is yet another exciting geometry problem-
> Q.In a triangle ABC, point P is taken so that, angle
> PAC=30 degree, angle PCA=10 degree, angle PCB=20
> degree and angle PAB=100 degree.Find angle ABP.




Here is the solution-
Q.In a triangle ABC, point P is taken so that, angle PAC=30 degree, angle PCA=10 degree, angle PCB=20 degree and angle PAB=100 degree.Find angle ABP.
Sol.-Contruct an equilateral triangle MAC so that M is on the same side of AC as B.Join BM.
Considering triangles PAM and PAC.
Angle PAM=angle PAC=30 degree.
PA is common side and AM=AC.
So, triangle PAM is congruent to triangle PAC.
Angle PAM=angle PCA=10 degree.
Considering triangles BCA and BCM.
Angle ACB=angle MCB=30 degree.
BC is common side and MC=AC.
So, triangle BCA is congruent to triangle BCM.
Angle CAB=angle CMB=130 degree.
Angle BMP=angle BMC- angle PMC=130-50=80 degree.
Angle PAB=100 degree.
Considering quadrilateral ABMP,
Angle BMP+angle PAB=100+80=180 degree.
So, quadrilateral ABMP is cyclic.
Hence,angle ABP=angle AMP=10 degree.



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