please help me with no.6
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since
AB=AC
then
angle ABC = angle ACB, (since angle opposite to equal sides are equal),
let angle ABC=angle ACB=x,
now in ∆ABC, by angle sum property
40°+angle ABC + angle ACB=180°,
angleABC+angleACB=180°-40°,
x+x=140°,
then
x=70°
now
angle ABC=angle ABD + angle DBC,
70°=30°+angle DBC,
then
angle DBC=40°,
now
angle BDC=angle ABD+angle BAD, (by exterior angle property),
then
angle BDC=30° + 40°=70°
now in ∆BDC, using angle sum property,
angle BDC+angle BCD+angle DBC=180°,
70°+angle BCD+40°=180°,
angle BCD=180°-70°-40°,
angle BCD=180°-110°,
angle BCD=70°,
since
angle BDC=angle BCD=70°,
then
BD=BC (since sides opposite to equal angles are equal)
AB=AC
then
angle ABC = angle ACB, (since angle opposite to equal sides are equal),
let angle ABC=angle ACB=x,
now in ∆ABC, by angle sum property
40°+angle ABC + angle ACB=180°,
angleABC+angleACB=180°-40°,
x+x=140°,
then
x=70°
now
angle ABC=angle ABD + angle DBC,
70°=30°+angle DBC,
then
angle DBC=40°,
now
angle BDC=angle ABD+angle BAD, (by exterior angle property),
then
angle BDC=30° + 40°=70°
now in ∆BDC, using angle sum property,
angle BDC+angle BCD+angle DBC=180°,
70°+angle BCD+40°=180°,
angle BCD=180°-70°-40°,
angle BCD=180°-110°,
angle BCD=70°,
since
angle BDC=angle BCD=70°,
then
BD=BC (since sides opposite to equal angles are equal)
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