Solve question no. 2 please...
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ADC +ABC =180
ABC = 50
since , perpendicular bisector of a chord always passes through the centre
Let AB intersects EC at M
In ∆EMB and ∆CMB, we have
EM =CM( AB is the perpendicular bisector)
BM=BM (common).
EMB =CMB=90
Thus , ∆EMB ~=∆CMB
→CBM = EBM =50 (cpct)
→CBE = 100
ABC = 50
since , perpendicular bisector of a chord always passes through the centre
Let AB intersects EC at M
In ∆EMB and ∆CMB, we have
EM =CM( AB is the perpendicular bisector)
BM=BM (common).
EMB =CMB=90
Thus , ∆EMB ~=∆CMB
→CBM = EBM =50 (cpct)
→CBE = 100
lala03:
for what??
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