Question

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Answer 1

$\huge{\mathcal{\underline{\green{AnSWer}}}}$

Given triangle ABC angle C is right angle

Draw CD perpendicular on line AB

In ΔACB and ΔBDC

∠ACB=∠BDC

∠B=∠B

∴ΔACB∼ΔBDC

∴$\frac{BC}{BD} = \frac{AC}{CD} = \frac{AB}{BC} \\ \\ \\ </p><p></p><p></p><p>∴BC {}^{2} =BD×AB - - - - - (1)  \\ \\ </p><p>$

ΔACB and ΔADC

∠ACB=∠ADC

∠A=∠A

∴ΔACB∼ΔADC

∴$\frac{BC}{CD} = \frac{AC}{AD} = \frac{AB}{AC} \\ \\ AC {}^{2} =AB×AD     - - - - (2)$

Divide (1) by (2) we get

∴$\frac{BC {}^{2} }{AC {}^{2} } = \frac{BD×AB}{AB×AD} = \frac{BD}{AD}$

Hence proved

Answer 2

Answer:

Given triangle ABC angle C is right angle

Draw CD perpendicular on line AB

In ΔACB and ΔBDC

∠ACB=∠BDC

∠B=∠B

∴ΔACB∼ΔBDC

ΔACB and ΔADC

∠ACB=∠ADC

∠A=∠A

∴ΔACB∼ΔADC

∆ACB ~ ∆ADC

HENCE PROVED .