Question

In the triangle ABC BAC=90 and AD perpendicular to BC Prove that AC2 =BC.DC​

Answer 1

to prove:AC2=BC×DC

AC×AC=BC×DC

AC÷DC=BC÷AC

proof:In triangle ADC n BAC

angleA=angleD(90°)

angleA=angleA(common angle)

triangle ADC is similar to BAC

corresponding sides r in same ratio

BC÷AC=AC÷DC

AC2=BC×DC