Question

Prove that AB2+AC2=(BD+CD)2
${ab}^{2} + {ac}^{2} = {(bc + cd)}^{2}$
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Answer 1

Answer:

$AB^2+AC^2=(BD+CD)^2$

Step-by-step explanation:

concept used:

Pythagors theorem:

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

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

Given: AD ⊥ BC

Also

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

In ΔADC,

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

In ΔADB,

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

Adding (2) and (3) we get

$AB^2+AC^2\\\\= AD^2+BD^2+AD^2+CD^2\\\\= BD^2+CD^2+2AD^2\\\\= BD^2+CD^2+2*BD*CD\:\:(using(1))\\\\=(BD+CD)^2$