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given:AB=AC
BD=BC
to prove: angle BAD=angleCAD
angleBDA=angleCDA
proof:in ∆ABD & ∆ACD
I)AB=AC(given)
ii)BD=BC(given)
iii)AD=AD(common)
∆ABDis congruent to ∆ACD(by SSS cong.rule)
angleBAD=angle CAD
[by cpct]
angleBDA=angleCDA
[by cpct]...........hope this helps you....
BD=BC
to prove: angle BAD=angleCAD
angleBDA=angleCDA
proof:in ∆ABD & ∆ACD
I)AB=AC(given)
ii)BD=BC(given)
iii)AD=AD(common)
∆ABDis congruent to ∆ACD(by SSS cong.rule)
angleBAD=angle CAD
[by cpct]
angleBDA=angleCDA
[by cpct]...........hope this helps you....
aryan1191:
wlcm
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