Ad andBE are the altitude of anisosceles triangle ABC with AC =BC prove that AE=BD
Answer :
Given AD and BE are altitude and AC = BC , So we have our figure , As :
Here
BEA = BEC = 90 --------------- ( 1 )
And
ADB = ADC = 90 ---------------- ( 2 )
So from equation 1 and 2 , we can say
BEA = ADB = 90 ---------------- ( 3 )
And As given ABC is a isosceles triangles so , from base angle theorem ,we can say that
CAB = CBA ----------------- ( 4 )
Now In BAE and ABD
BEA = ADB ( From equation 3 )
EAB = DBA ( As CAB = EAB ( same angles ) And CBA = DBA ( same angles ) And from equation 4 we know CAB = CBA )
And
AB = AB ( Common side )
Hence
BAE ABD ( By AAS rule )
So,
AE = BD ( By CPCT rule ) ( Hence proved )
Given AD and BE are altitude and AC = BC , So we have our figure , As :
Here
BEA = BEC = 90 --------------- ( 1 )
And
ADB = ADC = 90 ---------------- ( 2 )
So from equation 1 and 2 , we can say
BEA = ADB = 90 ---------------- ( 3 )
And As given ABC is a isosceles triangles so , from base angle theorem ,we can say that
CAB = CBA ----------------- ( 4 )
Now In BAE and ABD
BEA = ADB ( From equation 3 )
EAB = DBA ( As CAB = EAB ( same angles ) And CBA = DBA ( same angles ) And from equation 4 we know CAB = CBA )
And
AB = AB ( Common side )
Hence
BAE ABD ( By AAS rule )
So,
AE = BD ( By CPCT rule ) ( Hence proved )