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Question

Question 14
The perpendicular from A on side BC of a ΔABC intersects BC at D such that DB = 3CD .Prove that 2AB2=2AC2+BC2.

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Solution

Given that in ΔABC, we have
ADBC and BD=3CD
In right angle triangles ADB and ADC, we have
AB2=AD2+BD2...(i)
AC2=AD2+DC2...(ii) [By Pythagoras theorem]
Subtracting equation (ii) from equation (i), we get
AB2AC2=BD2DC2
=9CD2CD2[BD=3CD]
=8CD2
=8(BC4)2[Since,BC=DB+CD=3CD+CD=4CD]
Therefore, AB2AC2=BC22
2(AB2AC2)=BC2
2AB22AC2=BC2
2AB2=2AC2+BC2

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