English, asked by nirupamdas12282, 10 months ago

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

Attachments:

Answers

Answered by prajkumaripatel
5

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

Answered by TRISHNADEVI
4

   \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{ =&gt;  \frac{1}{2}  \angle \: PRS = \: \frac{1}{2}  \angle \: P + \: \frac{1}{2}  \angle \: PQR}}} \\  \\   \bf{ \sf{ \red { =&gt;}}}  \underline{ \bf{ \sf{ \green{  \:  \:  \: TRS = \: \frac{1}{2}  \angle \: P + TQR \:  \: }}}}  \:  \:  \:  -  -  -  -  -  &gt; (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 \:  \:  }}}} \:  \:  \:  -  -  -  -  -  &gt; (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{ =&gt; \frac{1}{2} \angle \: P = \angle \:T}}} \\  \\  \bf{ \sf{ \red{ =&gt;  \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>

Similar questions