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Re: Yet another exciting geometry problem
Posted:
Dec 19, 2010 10:54 AM


> 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=13050=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.



