Question

in the given figure PQR is a triangle where PS, PQ and RS are the bisectors of angleP, angle Q and angle R respectively.

1) if angle PRQ is greater than angle PQR,prove that SQ greater than SR.

2)if angle PRQ=110°and angle PQR=40,prove that SPis greater than SQ​

Answer 1

1.  In ΔPQR

   Given that

   ∠PRQ >∠PQR     ...1)

   Divide both side by 2

   $\frac{\angle PRQ}{2}> \frac{\angle PQR}{2}$

  ∠SRQ >∠SQR      (because line SR and SQ bisect the ∠R and ∠Q)

   So

   SQ>SR  (because side opposite greater angle are greater)

2.  In ΔPQR

   Given that

   ∠PRQ =110°  

   ∠PQR =40°     ...2)

   So

   ∠PRQ+∠PQR+∠RPQ = 180°  (from triangle property)

    110°+40°+∠RPQ =180°

    So

    ∠RPQ =30°     ...3)

3. From equation 2) and equation 3)

   ∠PQR >∠RPQ      (because ∠PQR =40° and ∠RPQ =30°)

    Divide both side by 2

    $\frac{\angle PQR}{2}> \frac{\angle RPQ}{2}$

    ∠PQS >∠SPQ    (because line SP and SQ bisect the ∠P and ∠Q)

    So

    SP > SQ  (because side opposite greater angle are greater)

   

   

 

Answer 2

Answer:

the given figure PQR is a triangle where PS, PQ and RS are the bisectors of angleP, angle Q and angle R respectively.

1) if angle PRQ is greater than angle PQR,prove that SQ greater than SR.

2)if angle PRQ=110°and angle PQR=40,prove that SPis greater than SQ