In triangle ABC, AB=AC and D is a point on side AC such that BC2=AC*CD. Prove that BD=BC
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Prove the smaller and the bigger triangles similar, you'll surely land-up with the answer!
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Hey ! Frnd . your answer is below. I hope it helps.
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BC2= AC.CD
BC.BC=AC.CD
BC/CD=AC/BC
ALSO C= C ( COMMON ANGLES)
∆ACB ~ ∆ BCD ( SAS SIMILARITY
AC/BC= AB/BD
AC/BC=AC/BD (AS AB = AC)
1/BC=1/BD
BC=BD
BC.BC=AC.CD
BC/CD=AC/BC
ALSO C= C ( COMMON ANGLES)
∆ACB ~ ∆ BCD ( SAS SIMILARITY
AC/BC= AB/BD
AC/BC=AC/BD (AS AB = AC)
1/BC=1/BD
BC=BD
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