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In DeltaABC,angleACB=90^(@) seg CD b...

In `DeltaABC,angleACB=90^(@)`
seg CD `bot` seg AB
seg DE `bot` seg CB.
Show that: `CD^(2)xxAC=ADxxABxxDE`

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In ` Delta ACB , angle ACB = 90^(@)`
seg `CD bot ` hypotenuse AB …(Given)
` :. ` by therorem of geometric mean
` CD^(2) = AD xx DB" "` …(i)
In ` Delta ACB and Delta DEB," "`
In `Delta ACB and Delta DEB,`
` angle ACB cong angle DEB " ...(Each measures " (90^(@))`
` :. Delta ACB ~ DeltaDEB " "` ...(A A test of similarity )
`(AC)/(DE) = (AB)/(DB)" "` ...( Corresponding sides of similar Deltas )
` :. AC = (AB)/(DB) xx DE`
Multiplying (1) and (2)
` CD^(2) xxAC = AD xxDB xx (AB)/(DB) xxDE`
` :. CD^(2) xx AC = AD xx AB xx DE`
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