In the figure △ABC is a right angled triangle with right angle at B. BD is perpendicular to AC. Then which of the following options will hold true?
AB2 = AD × DC
AB2 = AD × AC
AD2 = DC × AC
AB2 = DC2 + AD2
In △ABC and △ADB ∠ABC=∠ADB=90o ∠A=∠A (common angle) Therefore, △ABC∼△ADB by AA similarity ABAD=ACAB AB2=AC∗AD
In △ABC , B is the right angle and BD is perpendicular to AC, then:
△ABC is a right angled triangle, right angled at B. BD is a perpendicular as shown. Which of the following is true?
ΔABD is a right triangle right-angled at A and AC \perp BD. Show that (i) AB2=BC.BD (ii) AC2=BC.DC (iii) AD2=BD.CD (iv) AB2AC2=BDDC