Question

In ∆ABC, AD perpendicular to BC
T.P. T : AB²+ CD^2= AC² + DB²


Proof this​

Answer 1

Step-by-step explanation:

In triangle ADB,

AB² = AD²+ BD² ...(Pyrhagorus Theorem)

AD²=AB²-BD² ....(1)

In triangle ADC,

AC² = AD² + CD² ...(Pythagorus Theorem)

AD² = AC² - CD² ....(2)

From (1) and (2),

AC² - CD² = AB² - BD²

AB²+ CD² = AC² + BD² , which is the required result.