Question

The perpendicular from a on side bc of a triangle abc intersect bc at d such that bd equal to 3 cd

Answer 1

I know your question is not complete that's why I send u ans and question both its little blur but it's right ans and I think it's help u

Answer 2

Answer:

Given that in ΔABC, we have

AD ⊥BC and BD = 3CD

In right angle triangles ADB and ADC, we have

AB² = AD²+ BD² ...(i)

AC² = AD² + DC²...(ii) [By Pythagoras theorem]

Subtracting equation (ii) from equation (i), we get

AB²- AC² = BD² - DC²

= 9CD² - CD² [∴ BD = 3CD]

= 9CD² = 8(BC/4)² [Since, BC = DB + CD = 3CD + CD = 4CD]

Therefore, AB² - AC² = BC²/2

⇒ 2(AB² - AC²) = BC²

⇒ 2AB² - 2AC²= BC²

∴ 2AB² = 2AC²+ BC².

please follow me and mark as brainliest...