Question

In triangle PQR, angel PQR =90┬░, seg QS perpendicular seg QR, seg QM is angle bisector of angle PQR. Prove that Pm square/ MR square= PS / SR

Answer 1

Answer:

In figure angle PQR = 90°, seg QN ⊥ seg PR , PR = 9, NR = 16.Find QN

Answer 2

Step-by-step explanation:

Given:

In ΔPQR ,

$\angle PQR=90^{0}\\QS\perp PR\\QM\ is\ the\ angle\ bisector\ of\ \angle PQR$

To Prove: $PM^{2}/MR^{2}=PS/SR$

Calculation:

$In\ \bigtriangleup PQR: QM\ is\ angle\ bisector\ of\ \angle PQR\\PM/MR=PQ/QR(Property\ of\ angle\ bisector\ of\ triangle)\\\\PM^{2}/MR^{2}=PQ^{2}/QR^{2}-(equation1)\\\\In\ \bigtriangleup \ PQR-QS\perp PR\\\\\bigtriangleup PQR\cong \bigtriangleup RSQ\cong \bigtriangleup PSQ$

$\bigtriangleup PSQ\cong \bigtriangleup PQR\\\\\therefore PQ^{2} =PR\times PS-(equation-2)\\\\\bigtriangleup RSQ\cong \bigtriangleup PQR\\\\\therefore QR^{2}=PR\times SR-(equation-3)\\\\from\ equation-1, 2, 3:\\PM^{2}/MR^{2}=PR\times PS/PR\times SR\\PM^{2}/MR^{2}=PS/SR$

Hence Proved.