In Fig. 6.17, Q > R, PA is the bisector of QPR and PM QR. Prove that APM = (Q – R).


It is given to us –


Q > R


PA is the bisector of QPR


QPA = RPA - - - - (i)


PM QR


PMR = PMQ = 90° - - - - (ii)


We have to prove that APM = 1/2 (Q - R)


Since, the sum of the three angles of a triangle is equal to 180°, in ΔPQM,


PQM + PMQ + QPM = 180°


PQM + 90° + QPM = 180° [From equation (ii)]


PQM + QPM = 180° - 90°


PQM + QPM = 90°


PQM = 90° - QPM - - - - (iii)


Similarly, in ΔPMR, the sum of the three angles of a triangle is equal to 180°


PMR + PRM + RPM = 180°


90° + PRM + RPM = 180° [From equation (ii)]


PRM + RPM = 180° - 90°


PRM + RPM = 90°


PRM = 90° - RPM - - - - (iv)


Now, on subtracting equation (iv) from equation (iii), we get


PQM - PRM = (90° - QPM) - (90° - RPM) ×


Q - R = 90° - QPM - 90° + RPM


Q - R = RPM - QPM


Q - R = (RPA + APM) - (QPA - APM)


Q - R = RPA + APM - QPA + APM


Q - R = QPA + 2 × APM - QPA [From equation (i)]


Q - R = 2 × APM


APM = 1/2 (Q - R)


Hence, proved.


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