Question

D is the point on the side BC of a triangle ABC such that angle ADC = angle BAC. Show that
${ca}^{2} = cb \times cd$



​

Answer 1

Answer:

Hey mate following is your ans

/|BAC nd /|ADC

/BAC=/ADC {given}

/C=/C. {common angle}

/|BAC similar /|ADC {AA}

ba/ad=ac/dc=bc/ac ( sides of similar tri)

ac/dc=bc/ac

ac^2=bc ×dc