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# In figure below, D is the mid-point of side BC and $AE \perp BC$. If $B C=a, A C=b, A B=C, E D=x, A D=p$ and $A E=h,$ prove that $b^{2}+c^{2}=2 p^{2}+\frac{a^{2}}{2}$."

Given:

In the given figure, D is the mid-point of side BC and $AE \perp BC$.

$B C=a, A C=b, A B=C, E D=x, A D=p$ and $A E=h$.

To do:

We have to prove that $b^{2}+c^{2}=2 p^{2}+\frac{a^{2}}{2}$.

Solution:

In $\triangle AED$, by using Pythagoras theorem,

$AD^2=AE^2+ED^2$

$AE^2=AD^2-ED^2$.....(i)

In $\triangle AEC$, by using Pythagoras theorem,

$AC^2=AE^2+EC^2$ $b^2=(AD^2-ED^2)+(ED+DC)^2$     (From (i))

$b^2=AD^2-ED^2+ED^2+DC^2+2ED\times DC$

$b^2=AD^2+DC^2+2DC\times ED$

$b^2=p^2+(\frac{a}{2})^2+2\times(\frac{a}{2})\times x$   (Since $DC=\frac{BC}{2}$)

$b^2=p^2+\frac{a^2}{4}+ax$......(ii)

In $\triangle AEB$, by using Pythagoras theorem,

$AB^2=AE^2+BE^2$

$c^2=(AD^2-ED^2)+(BD-ED)^2$     (From (i) and $BE=BD-ED$)

$c^2=AD^2-ED^2+BD^2+ED^2-2BD\times ED$

$c^2=AD^2+BD^2-2BD\times ED$

$c^2=p^2+(\frac{a}{2})^2-2\times(\frac{a}{2})\times x$   (Since $DC=\frac{BC}{2}$)

$c^2=p^2+\frac{a^2}{4}-ax$.....(iii)

Adding equations (ii) and (iii), we get,

$b^2+c^2=p^2+\frac{a^2}{4}+ax+p^2+\frac{a^2}{4}-ax$

$b^2+c^2=2p^2+2\times\frac{a^2}{4}$

$b^2+c^2=2p^2+\frac{a^2}{2}$

Hence proved.

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Updated on: 10-Oct-2022

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