Question

In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

(i) `"AC"^2 = "AD"^2 + "BC"."DM" + (("BC")/2)^2`

(ii) `"AB"^2 = "AD"^2 - "BC"."DM" + (("BC")/2)^2`

(iii) `"AC"^2 + "AB"^2 = 2"AD"^2 + 1/2"BC"^2`

Answer 1

(i) Applying Pythagoras theorem in ΔAMD, we obtain

AM² + MD² = AD² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC²

AM² + (MD + DC)² = AC²

(AM² + MD²) + DC² + 2MD.DC = AC²

AD² + DC² + 2MD.DC = AC² [Using equation (1)]

Using the result, DC = `"BC"/2`, we obtain

`"AD"^2+(("BC")/2)^2 + 2"MD".(("BC")/2) = "AC"^2`

`"AD"^2+(("BC")/2)^2 + "MC" xx "BC" = "AC"^2`

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AB² = AM² + MB²

= (AD² − DM²) + MB²

= (AD² − DM²) + (BD − MD)²

= AD² − DM² + BD² + MD² − 2BD × MD

= AD² + BD² − 2BD × MD

= `"AD"^2+(("BC")/2)^2 - 2(("BC")/2) xx "MD"`

= `"AD"^2 + ("BC"/2)^2 - "BC" xx "MD"`

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AM² + MB² = AB² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC² … (2)

Adding equations (1) and (2), we obtain

2AM² + MB² + MC² = AB² + AC²

2AM² + (BD − DM)² + (MD + DC)² = AB² + AC²

2AM²+BD² + DM² − 2BD.DM + MD² + DC² + 2MD.DC = AB² + AC²

2AM² + 2MD² + BD² + DC² + 2MD (− BD + DC)

= AB² + AC²

= `2("AM"^2 + "MD"^2) + (("BC")/2)^2 + (("BC")/2)^2 + 2"MD" ((-"BC")/2 + ("BC")/2) = "AB"^2 + "AC"^2`

`"2AD"^2 + ("BC"^2)/2 = "AB"^2 + "AC"^2`