If the bisectors of angles ∠PQR and ∠PRQ of triangle PQR meet at a point M, then prove that angle QMR = 90+1/2 angle P
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Given : bisectors of angles ∠PQR and ∠PRQ of triangle PQR meet at a point M
To find : prove that angle QMR = 90+1/2 angle P
Solution:
in Δ PQR
∠PQR + ∠PRQ + ∠P = 180° ( sum of angles of triangle)
=>∠PQR + ∠PRQ = 180° - ∠P
in Δ MQR
∠MQR + ∠MRQ + ∠QMR = 180° ( sum of angles of triangle)
∠MQR = ∠PQR/2
∠MRQ = ∠PRQ/2
=> ∠PQR/2 + ∠PRQ/2 + ∠QMR = 180°
=> (∠PQR + ∠PRQ )/2 + ∠QMR = 180°
=> (180° - ∠P)/2 + ∠QMR = 180°
=> 90° - ∠P/2 + ∠QMR = 180°
=> ∠QMR = 90° + ∠P/2
QED
Hence proved
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