D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC.Prove that:CA²=CBxCD.
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ABC is a triangle in which AB=AC and D is a point on AC such that BC^2 = AC ×CD. Prove that BD= BC.
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Given in ΔABC, ∠ADC = ∠BAC
Consider ΔBAC and ΔADC
∠ADC = ∠BAC (Given)
∠C = ∠C (Common angle)
∴ ΔBAC ~ ΔADC (AA similarity criterion)
=AB/AD= CB/CA= CA/CD
consider,CB/CA=CA/CD
=CA^2=CB.CD
hope it's clear to you...
Consider ΔBAC and ΔADC
∠ADC = ∠BAC (Given)
∠C = ∠C (Common angle)
∴ ΔBAC ~ ΔADC (AA similarity criterion)
=AB/AD= CB/CA= CA/CD
consider,CB/CA=CA/CD
=CA^2=CB.CD
hope it's clear to you...
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