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Prove that the circle drawn on any one o...

Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

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Given : `A` `Delta ABC` in which `AB` `=` `AC` and a circle is drawn with `AB` as diameter, intersecting `BC` at `D`.
To Prove : ` BD` `=` `CD.`
Construction : Join `AD`.

Proof : We known that an angle in a semicircle is a right angle.
`:. /_ ADB = 90^(@)` ....(i)
Also, DBC being a straight line, we have
`/_ DDB + /_ADC = 180^(@)`
`implies 90^(@) +/_ADC = 180^(@)` [ using (i) ]
`implies /_ ADC = 90^(@)` ....(i)
Now, in `Delta `ADB and `Delta ADC`, we have
AB `=` AC [given]
`/_ ADB = /_ ADC` [each equal to `90^(@)]`
`AD ` `=` `AD` [common ]
`:. Delta ~= Delta ADC `[ by `RHS` - congurence]
Hence , `BD = CD`
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