Question

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Answer 1

Step-by-step explanation:

Have Given :-

OQ and OR are internal bisector of ∠Q and ∠R

PQR is a triangle

also from figure OQR is a triangle

Let ∠Q = 2x

and ∠R = 2y

Now in ∆PQR

$∠Q + ∠R + ∠P = 180°$

$2x +2y + ∠P = 180°$

$2(x +y ) = 180° - ∠P$

$x +y = \frac{180° - ∠P}{2}$

$x + y = \frac{ \cancel{180°}}{ \cancel2} - \frac{ ∠P}{2}$

$x + y = 90° - \frac{1}{2} ∠P - - - (i)$

Now in ∆OQR

$\angle RQO + \angle QRO+ \angle QOR = 180°$

$Let \: \angle RQO = x$

$and \: \angle QRO = y$

$x + y + \angle{QOR }= 180°$

$\angle{QOR}= 180° - (x + y) - - - (ii)$

Now from Equation (i) and (ii) -

$\angle{QOR} = 180° - (90° - \frac{1}{2} \angle{P})$

$\angle{QOR} = 180° - 90° + \frac{1}{2} \angle{P}$

$\angle{QOR} = 90° + \frac{1}{2} \angle{P}$

Hence proved

Answer 2

Answer:

Answer the above attachment