document.write( "Question 61092: In triangle ABC , AB=AC. If there is a point P strictly between A and B such that AP=PC=CB, then angle A = ? \n" ); document.write( "
Algebra.Com's Answer #42017 by venugopalramana(3286)\"\" \"About 
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In triangle ABC , AB=AC.
\n" ); document.write( "HENCE
\n" ); document.write( "ANGLE B = ANGLE C = X SAY\r
\n" ); document.write( "\n" ); document.write( "If there is a point P strictly between A and B such that AP=PC=CB,
\n" ); document.write( "IN TRIANGLE PCB , PC=CB....HENCE
\n" ); document.write( "ANGLE B = X = ANGLE BPC......................1
\n" ); document.write( "IN TRIANGLE PAC , AP=PC....HENCE
\n" ); document.write( "ANGLE A = Y SAY = ANGLE PCA
\n" ); document.write( "ANGLE C = X = ANGLE BCP + ANGLE PCA = ANGLE BCP + Y
\n" ); document.write( "ANGLE BCP = X-Y
\n" ); document.write( "ANGLE BPC = EXTERNAL ANGLE AT P FOR TRIANGLE APC
\n" ); document.write( "= SUM OF OPPOSITE INTERIOR ANGLES = ANGLE A + ANLE PCA = Y+Y = 2Y........2
\n" ); document.write( "FROM EQN.1 AND EQN.2.....X=2Y
\n" ); document.write( "SUM OF 3 ANGLES IN TRINGLE PBC =180 = ANGLE PBC + ANGLE BCP + ANGLE BPC
\n" ); document.write( "= X + X-Y+2Y =180
\n" ); document.write( "2X+Y=180
\n" ); document.write( "2*2Y+Y=180
\n" ); document.write( "5Y=180
\n" ); document.write( "Y=36
\n" ); document.write( "HENCE ANGLE A =36\r
\n" ); document.write( "\n" ); document.write( " then angle A = ? = 36
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