# In Fig. 6.17, <sp

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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