13] AABC is right angled at C. From any point D
on side AB, DH 1 AC, DK 1 BC.
Show that AD. BD = AH.HC + BK. CK.
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Answer:
Consider △ACD&△ABC
∠CAD=∠CAB
∠CDA=∠ACB
∴△ACD∼△ABC,{ By AA similarty criterion}
⟹ABAC=ACAD
∴AC2=AB×AD−(1)
Similarily,△BCD∼△BAC{ by AA similarity criterion}
⟹BABC=BCBD
∴BC2=BA×BD−(2)
∴AC2BC2=AB×ADAB×BD
∴BC2×AD=AC2×BD
Step-by-step explanation:
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