AD=AE,BD=EC.Prove that triangle ABC is an isosceles triangle
Answers
Answered by
12
AD=AE ........(1)
BD=EC ........(2)
ON ADDING BOTH EQUATION 1 ,2
AD+BD=AE+EC
AB=AC
TWO SIDES OF THIS ∆ABC are EQUAL
SO ∆ABC is isosceles
BD=EC ........(2)
ON ADDING BOTH EQUATION 1 ,2
AD+BD=AE+EC
AB=AC
TWO SIDES OF THIS ∆ABC are EQUAL
SO ∆ABC is isosceles
Answered by
2
In isosceles ∆, only two sides are equal third side is not equal....
Let, A ∆ABC in which two sides are equal , AD = AE, BD = EC.
A/q,. there are two triangles first is ∆ADE & ∆BDC, we have
prove that: AD = BC (sides of ∆ADE & ∆BDC)
angle D = angle D ( opp. angle of both ∆)
DE = DC ( side of both ∆ADE & ∆BDC)
angle E = angle C (opp. angle of both ∆)
so,∆ADE =~ ∆BDC ( By sss rule )
DE= DC. ( By C.P.C.T.)
So , AD = AE & BD = EC are two isosceles ∆.
Let, A ∆ABC in which two sides are equal , AD = AE, BD = EC.
A/q,. there are two triangles first is ∆ADE & ∆BDC, we have
prove that: AD = BC (sides of ∆ADE & ∆BDC)
angle D = angle D ( opp. angle of both ∆)
DE = DC ( side of both ∆ADE & ∆BDC)
angle E = angle C (opp. angle of both ∆)
so,∆ADE =~ ∆BDC ( By sss rule )
DE= DC. ( By C.P.C.T.)
So , AD = AE & BD = EC are two isosceles ∆.
Similar questions