in triangle abc angle abc =90 and bm is the altitude, if am = 16 mc prove that ab = 4bc
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Given :
∠ABC=90°
AM=16MC
To proof : AB=4BC
let MC=x
∴AM=16x
In △AMB & △ABC
∠A=∠A (common)
∠AMB=∠ABC (90°)
△AMB∼△ABC (by AA)
∴
AB
AM
=
BC
BM
⟹AM×BC=AB×BM⟶(1)
Now in △CMB & △CBA
∠C=∠C (common)
∠CMB=∠CBA (90°)
△CMB∼△CBA (by AA)
∴
CB
CM
=
AB
MB
⟹AB×CM=MB×BC⟶(2)
Divide (1) by (2)
AB×CM
AM×BC
=
MB×BC
AB×BM
[∵AM=16CM]
AB×CM
16×CM×BC
=
BC
AB
⟹16BC
2
=AB
2
AB=4BC
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