can you solve at least any one of them
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Here's the 2nd ques solution :-
To prove :- DE . BD = DC . BE
Proof,
In ΔCDB and ΔDEB
angle B = angle B
angle BDC = angle BED
ΔCDB ~ ΔDEB ( by AA)
So, DC/BD = DE/BE
DE . BD = DC . BE
To prove :- DE . BD = DC . BE
Proof,
In ΔCDB and ΔDEB
angle B = angle B
angle BDC = angle BED
ΔCDB ~ ΔDEB ( by AA)
So, DC/BD = DE/BE
DE . BD = DC . BE
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