`Delta ABC` में, भुजा `AB` को `D` तक इस प्रकार बढ़ाया गया है कि `BD = BC` यदि `angle B = 60^(@)` तथा ` angleA = 70^(@)` तो सिद्ध कीजिए कि -
(i) `AD gt CD " "(ii) AD gt AC`

Explanation

`angleA + angleB + angleC = 180^(@)`
`:. " " angleC = 180^(@) - angleA - angleB`
` = 180^(@) - 70^(@) - 60^(@) = 50^(@)`
`angleCBD = angleA + angleC`
` = 70^(@) + 50 ^(@) = 120^(@)`
(बहिष्कोण प्रमेय )
`Delta CBD " में , " angleCBD + angleBCD + angleCDB= 180^(@)`
`rArr 120^(@) + angleBCD + angleCDB = 180^(@)`
या `angleBCD + angleCDB = 60^(@)`
लेकिन `angleBCD = angleCDB " "[:' BD = BC]`
`:. angleBCD = angleCDB =30^(@)`
अब `angle ACD = angleC + angleBCD`
` = 50^(@) + 30^(@) = 80^(@)`
`Delta ACD" में , " angleACD gt angleCAD`
` AD gt CD`
तथा `angleACD (=80^(@)) gt angleCDA (=30^(@))`
अतः `AD gt AC`