Solve the following questions and explain properly-
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Hey!
solve in fig..
Hope it helps you
solve in fig..
Hope it helps you
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Abhiranjanraj:
wellcome
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In triangleAPC and trianglePBD
anglePAC= anglePBD (given)
AP =PB (given)
angleAPC =angle BPD (vetically opposite angle)
therefore triangleAPC= trianglePBD
In triangleBMC and triangle BLC
angleBMC= angleBLC=90degree
BM=LC ( AB=AC)
BC=BC (common)
THEREFORE triangle BMC and triangle LBC
BL=CM (C.P.C.T.)
anglePAC= anglePBD (given)
AP =PB (given)
angleAPC =angle BPD (vetically opposite angle)
therefore triangleAPC= trianglePBD
In triangleBMC and triangle BLC
angleBMC= angleBLC=90degree
BM=LC ( AB=AC)
BC=BC (common)
THEREFORE triangle BMC and triangle LBC
BL=CM (C.P.C.T.)
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