In △ABP and △CDP, Seg AC and Seg BD intersect each other at point P and AP/CP = BP/DP. By which test is △ABP ~ △CDP? *
options enmese ek answer ha
A-A test
A-A-A test
S-A-S test
S-S-S test
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IN figure it is given that segAC and segBDintersect each other in point P such that CPAP=DPBP
InΔAPBandΔDCPCPAP=DPBP(given)∠APB=∠DPC(verticallyoppositeangle)∴ΔABP∼ΔCDP(SAS)
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