Advertisement Remove all ads

The perpendicular from A on side BC of a Δ ABC meets BC at D such that DB = 3CD. Prove that 2AB2 = 2AC2 + BC2. - Mathematics

Sum

The perpendicular from A on side BC of a Δ ABC meets BC at D such that DB = 3CD. Prove that 2AB2 = 2AC+ BC2.

Advertisement Remove all ads

Solution

Applying Pythagoras theorem for ΔACD, we obtain 

AC2 = AD2 + DC2

AD = AC2 - DC2 ......(1)

Applying Pythagoras theorem in ΔABD, we obtain

AB2 = AD2 + DB2

AD2 = AB2 - DB2 .....(2)

From equation (1) and equation (2), we obtain

AC2 - DC2 = AB2 - DB2 .....(3)

It is given that 3DC = DB

∴ `"DC" = "BC"/4` and `"DB"=(3"BC")/ 4`

Putting these values in equation (3), we obtain

`"AC"^2 - ("BC"/4)^2 = "AB"^2 - ((3"BC")/4)^2`

`"AC"^2 - "BC"^2/16 = "AB"^2 - (9"BC"^2)/4`

16AC2 - BC2 = 16AB2 - 9BC2

16AB2 - 16AC2 = 8BC2

2AB2 = 2AC2 + BC2

Concept: Areas of Similar Triangles
  Is there an error in this question or solution?
Advertisement Remove all ads
Advertisement Remove all ads

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×