Ex 6.5, 14
The perpendicular from A on side BC of a ΔABC intersects BC at D such that DB = 3 CD (see figure). Prove that 2AB2 = 2AC2 + BC2
Given: ΔABC with AD ⊥ BC
Also DB = 3 CD
To prove: 2AB2 = 2AC2 + BC2
Proof:
Let BC = x
Using Pythagoras theorem
(Hypotenuse)2 = (Height)2 + (Base)2
Now in right ∆ 𝐴𝐷𝐶
AC2 = AD2 + DC2
AC2 = AD2 + (𝑥/4)^2
AC2 = AD2 + 𝑥2/16
Similarly in right ∆ 𝐴𝐷𝐵
AB2 = AD2 + DB2
AB2 = AD2 + ( 3𝑥/4 )2
AB2 = AD2 + 9𝑥2/16
Subtracting (2) from (1)
AC2 – AB2 = AD2 + 𝑥2/16 – (𝐴𝐷2+9𝑥2/16)
AC2 – AB2 = AD2 + 𝑥2/16−𝐴𝐷2−9𝑥2/16
AC2 – AB2 = AD2 – AD2 + 𝑥2/16 −9𝑥2/16
AC2 – AB2 = (𝑥2 − 9𝑥2)/16
AC2 – AB2 = −8𝑥2/16
AC2 – AB2 = −( 𝑥2)/2
2AC2 – 2AB2 = – x2
Putting BC = x
2AC2 – 2AB2 = – BC2
2AC2 + BC2 = 2AB2
2AB2 = 2AC2 + BC2
Hence proved

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.