Q12 of 56 Page 127

On drawing a perpendicular from vertex A of a ΔABC on its opposite side BD AD2 = BD × DC is obtained. Then prove that ABC is a right angled triangle.


Given,


AD2 = BD × DC


|(1)


∠ADB = 90°


In ΔADC & ΔADB,



∠ADB = ∠CDA = 90°


ΔADC~ΔADB by SAS Similarity Rule


⇒ ∠CAD = ∠ABD


⇒ ∠ACD = ∠BAD …(1)


In ΔADC,


∠CAD + ∠ACD + ∠ADC = 180°


⇒ ∠CAD + ∠ACD = 180° – 90° = 90°


From (1),


⇒ ∠CAD + ∠BAD = 90°


⇒ ∠BAC = 90°


Thus, ABC is right angled triangle.


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