5. Triangle ABC is right-angled at C and CD is perpendicular to AB. yo that
BCP X AD- AC BD
Answers
Answered by
7
Answer:
Consider △ACD&△ABC
∠CAD=∠CAB
∠CDA=∠ACB
∴△ACD∼△ABC,{ By AA similarty criterion}
⟹
AB
AC
=
AC
AD
∴AC
2
=AB×AD−(1)
Similarily,△BCD∼△BAC{ by AA similarity criterion}
⟹
BA
BC
=
BC
BD
∴BC
2
=BA×BD−(2)
∴
AC
2
BC
2
=
AB×AD
AB×BD
∴BC
2
×AD=AC
2
×BD
solution
Answered by
1
Step-by-step explanation:
Consider △ACD&△ABC
∠CAD=∠CAB
∠CDA=∠ACB
∴△ACD∼△ABC,{ By AA similarty criterion}
⟹
AB
AC
=
AC
AD
∴AC
2
=AB×AD−(1)
Similarily,△BCD∼△BAC{ by AA similarity criterion}
⟹
BA
BC
=
BC
BD
∴BC
2
=BA×BD−(2)
∴
AC
2
BC
2
=
AB×AD
AB×BD
∴BC
2
×AD=AC
2
×BD
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