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AB=AC(given )
AP=AQ(given )
AB+AQ=AC+AP( AQ is given equal to AP)
So, BQ=PC consider it as equation 1
now in ∆APB and ∆AQC
BQ=PC(using equation 1)
PAB= QAC(vertically opposite angles are equal )
BC=BC(common in both triangles )
so ,∆APB is congruent to ∆AQC by SAS criterion
so BP=QC(cpct)
AP=AQ(given )
AB+AQ=AC+AP( AQ is given equal to AP)
So, BQ=PC consider it as equation 1
now in ∆APB and ∆AQC
BQ=PC(using equation 1)
PAB= QAC(vertically opposite angles are equal )
BC=BC(common in both triangles )
so ,∆APB is congruent to ∆AQC by SAS criterion
so BP=QC(cpct)
Anonymous:
hope it helped
It did! A lot! Thanks❤
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