In a ΔABC, AD is a median and AL ⊥ BC.

Prove that:

(a) AC² = AD² + BC • DL +

(b) AB² = AD² - BC • DL +

(c) AC² + AB² = 2AD² +

Question

In a ΔABC, AD is a median and AL ⊥ BC. Prove that: (a) AC 2 = AD 2 + BC • DL + (b) AB 2 = AD 2 - BC • DL + (c) AC 2 + AB 2 = 2AD 2 +

Philoid · Accepted Answer

(a) In right triangle ALC

Using Pythagoras theorem, we have

AC² = AL² + LC²

⇒ AC² = AD² − DL² + (DL + DC) ²

….(1)

(b) In right triangle ALD

Using Pythagoras theorem, we have

AL² = AD² - LD²

Again, in ΔABL

Using Pythagoras theorem, we have

AB² = AL² + LB²

⇒ AB² = AD² –DL² + LB²

⇒ AB² = AD²–DL² + (BD – DL)²

……(2)

(c) Adding (1) and (2)

In a ΔABC, AD is a median and AL ⊥ BC.Prove that:(a) AC2 = AD2 + BC • DL + (b) AB2 = AD2 - BC • DL + (c) AC2 + AB2 = 2AD2 +