Sum

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

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#### Solution

Applying Pythagoras theorem for ΔACD, we obtain

AC^{2} = AD^{2} + DC^{2}

AD^{2 } = AC^{2 }- DC^{2} ......(1)

Applying Pythagoras theorem in ΔABD, we obtain

AB^{2} = AD^{2} + DB^{2}

AD^{2} = AB^{2} - DB^{2} .....(2)

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

AC^{2 }- DC^{2} = AB^{2} - DB^{2} .....(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`

16AC^{2 }- BC^{2} = 16AB^{2} - 9BC^{2}

16AB^{2} - 16AC^{2 }= 8BC^{2}

2AB^{2} = 2AC^{2} + BC^{2}

Concept: Areas of Similar Triangles

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