Show that angle TPS=half of the difference of angle Q and R |Qis less than R| Triangle|@FOCUS4MATHS

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In Fig. 6.17, ∠Q ∠R, PA is the bisector of ∠QPR and PM perpendicular QR Prove that APM=1/2 (∠Q -∠R) In Fig. 6.17, ∠Q ∠R, PA is the bisector of ∠QPR and PM perpendicular QR Prove that APM=1/2 (∠Q -∠R)
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In given figure, PS is bisector of ∠QPR, PT⊥RQ and Q greater than R, show that ∠TPS = (1/2)(∠Q - ∠R) In given figure, PS is bisector of ∠QPR, PT⊥RQ and Q greater than R, show that ∠TPS = (1/2)(∠Q - ∠R)