Answers

2014-06-27T19:02:03+05:30
Lets take ΔABC B=90 BD perpend. to AC
In ΔABC, ∧B=90°
∧C+∧A=90
let ∧A=x
Therefore, ∧C=90-∧A
⇒90-x
In ΔADB,
∧D=90
∧A+∧ABD=90
∧ABD=90-∧A
⇒90-x

In ΔADB & ΔBDC
∧D=∧D=90
∧ABD=∧C       (=90-x)
ΔADB≈ΔBDC   (AA≈)
 \frac{BD}{CD} =  \frac{AD}{BD}
BD²=AD×CD
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2014-06-27T21:30:09+05:30
Lets take ΔABC B=90 BD perpend. to AC
In ΔABC, ∧B=90°
∧C+∧A=90
let ∧A=x
Therefore, ∧C=90-∧A
⇒90-x
In ΔADB,
∧D=90
∧A+∧ABD=90
∧ABD=90-∧A
⇒90-x

In ΔADB & ΔBDC
∧D=∧D=90
∧ABD=∧C       (=90-x)
ΔADB≈ΔBDC   (AA≈)

BD²=AD×CD

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