Question

Prove that in a triangle Ab square plus ac square is equal to bd plus cd whole square

Answer 1

AB

2

+AC

2

=(BD+CD)

2

concept used:

Pythagors theorem:

In a right angled square on the hypotenuse is equal to sum of the squares on the other two sides.

(a+b)^2=a^2+b^2+2ab(a+b)

2

=a

2

+b

2

+2ab

Given: AD ⊥ BC

Also

AD^2=BD*CD.............(1)AD

2

=BD∗CD.............(1)

In ΔADC,

AC^2=AD^2+CD^2........(2)AC

2

=AD

2

+CD

2

........(2)

In ΔADB,

AB^2= AD^2+BD^2........(3)AB

2

=AD

2

+BD

2

........(3)

Adding (2) and (3) we get

\begin{lgathered}AB^2+AC^2\\\\= AD^2+BD^2+AD^2+CD^2\\\\= BD^2+CD^2+2AD^2\\\\= BD^2+CD^2+2*BD*CD\:\:(using(1))\\\\=(BD+CD)^2\end{lgathered}

AB

2

+AC

2

=AD

2

+BD

2

+AD

2

+CD

2

=BD

2

+CD

2

+2AD

2

=BD

2

+CD

2

+2∗BD∗CD(using(1))

=(BD+CD)

2

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