Question

please answer this question......fast please please ​

Answer 1

Answer:

ok i will

GIVEN BISECTOR OF ag.PQR and ag.PRS MEET AT POINT T

to prove

ag.QTR=1/2QPR

prove,

ag.TRS=ag.TQR+ag.QTR

ag.QTR=ag.TRS-ag.TQR...... (1)

ag.SRP=AG.QPR+AG.PQR

2 AG.TRS=AG.QPR+2AG.TQR

1/2AG.QPR=AG.TRS-AG.TQR....... (2)

FROM 1 AND 2

AG.QTR-AG.TQR=1/2 AG.QPR

HOPE IT HELP YOU

SORRY I AM LATE

here ag. means angle

Answer 2

$\huge{ \bold{ \underline{ \overline{ \mid{ \blue{ \: \: SOLUTION \: \: } \mid}}}}}$

$</p><p></p><p> \bf{ \sf{In \: \: \triangle \: PQR , \: side \: \: QR \: \: is \: \: produced \: \: to \: \: S \: .}} \\ \\ \bf{ \: \: \: \: \: \: \: \: So , \: by \: \: exterior \: \: angle \: \: property, } \\ \\ \: \: \: \: \: \: \: \: </p><p> \bf{ \sf{ \green{ \angle \: PRS = \angle \: P + \angle \: PQR }}} \\ \\ \bf{ \sf{ \green{ => \frac{1}{2} \angle \: PRS = \: \frac{1}{2} \angle \: P + \: \frac{1}{2} \angle \: PQR}}} \\ \\ \bf{ \sf{ \red { =>}}} \underline{ \bf{ \sf{ \green{ \: \: \: TRS = \: \frac{1}{2} \angle \: P + TQR \: \: }}}} \: \: \: - - - - - > (1) \: \: \: \: \: \: \: \: \: \\ \\ </p><p></p><p> \bf{[ As \: \: QT \: \: and \: \: RT \: \: are \: \: \: bisectors \: \: of \: } \\ \bf{ \angle \: PQR \: \: and \: \: \angle \: PRS \: \: respectively ]} \\ \\ \\ </p><p> \bf{ \sf{In \: \: \triangle \: QRT \: \: , side \: \: QR \: \: is \: \: produced \: \: to \: \: S \: .}} \\ \\ \bf{</p><p>So , by \: \: exterior \: \: angle \: \: property \: , } \\ \\ </p><p> \underline{\bf{ \sf{ \green{ \: \: \angle \: TRS = \: \angle \: T + \angle \: TQR \: \: }}}} \: \: \: - - - - - > (2) \\ \\ \bf{</p><p>From \: \: (1) \: \: and \: \: (2) , \: we \: \: get \: , } \\ \\ \bf{ \sf{ \red{</p><p></p><p> \frac{1}{2} \angle \:P +\angle \: TQR = \angle \:T + \angle \:TQR }}} \\ \\ \bf{ \sf{ \red{ => \frac{1}{2} \angle \: P = \angle \:T}}} \\ \\ \bf{ \sf{ \red{ => \frac{1}{2} \angle \:QPR =\angle \: QTR }}} \\ \\ \underline{\underline{\bf{ \pink{ \: \: \therefore \: \angle \:QTR = \frac{1}{2} \angle \:QPR \: \: }}}}</p><p></p><p></p><p>$