In the adjoining figure, ∆ABC is an isosceles triangle in which AB = AC and AD is the bisector of ∠A.
Prove that:
(i) ∆ADB ≡ ∆ADC
(ii) ∠B = ∠C
(iii) BD = DC
(iv) AD ⊥ BC
Dear Student,
Please find below the solution to the asked query:
i ) In ADB and ADC
AB = AC ( Given )
BAD = CAD ( As given AD is angle bisector of A )
AD = AD ( Common side )
So,
ADB ADC ( By SAS rule ) ( Hence proved )
ii ) B = C ( By CPCT ) ( Hence proved )
iii ) BD = CD ( By CPCT ) ( Hence proved )
iv ) ADB = ADC ( By CPCT ) --- ( 1 )
And
ADB + ADC = 180 ( Linear pair angles )
Now we substitute value from equation 1 we get :
ADB + ADB = 180 ,
2 ADB = 180 ,
ADB = 90 , From equation 1 we get :
ADB = ADC = 90 , From this relation we can say that :
AD BC ( Hence proved )
Hope this information will clear your doubts about topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards
Please find below the solution to the asked query:
i ) In ADB and ADC
AB = AC ( Given )
BAD = CAD ( As given AD is angle bisector of A )
AD = AD ( Common side )
So,
ADB ADC ( By SAS rule ) ( Hence proved )
ii ) B = C ( By CPCT ) ( Hence proved )
iii ) BD = CD ( By CPCT ) ( Hence proved )
iv ) ADB = ADC ( By CPCT ) --- ( 1 )
And
ADB + ADC = 180 ( Linear pair angles )
Now we substitute value from equation 1 we get :
ADB + ADB = 180 ,
2 ADB = 180 ,
ADB = 90 , From equation 1 we get :
ADB = ADC = 90 , From this relation we can say that :
AD BC ( Hence proved )
Hope this information will clear your doubts about topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards