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△ABC की भुजा BC पर AD रेखा इस प्रकार है कि ∠ADC = ∠BAC, तो दिखाएँ कि AC2 = BC x CD.

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△ABC की भुजा BC पर AD रेखा इस प्रकार है कि ∠ADC = ∠BAC

दिया है,

\(\angle \mathrm{ADC}=\angle \mathrm{BAC}\)

सिद्ध करना है : \(\mathrm{AC}^2=\mathrm{BC} \times \mathrm{CD}\)

प्रमाण : \(\triangle \mathrm{ABC}\) तथा \(\triangle \mathrm{ADC}\) में, 

\(\angle \mathrm{C}=\angle \mathrm{C}\) उभयनिष्ठ है। 

\(\angle \mathrm{BAC}=\angle \mathrm{ADC}\) दिया है। 

कोण-कोण उप-प्रमेय से, 

\(\triangle \mathrm{BAC} \sim \angle \mathrm{ADC}\) तो 

\(\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{AC}}{\mathrm{DC}}\) 

अत: \(\mathrm{AC}^2=\mathrm{BC} \times \mathrm{DC}\) सिद्ध हुआ।

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