Question

in triangle abc ad is perpendicular to BC and AD square is equal to BD into CD prove that a b square + AC square is equal to BD + CD whole square​

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

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

Answer 2

Answer:

Proved AB² + AC² = (BD + CD)²

Step-by-step explanation:

In triangle abc ad is perpendicular to BC and AD square is equal to BD into CD prove that a b square + AC square is equal to BD + CD whole square​

AD ⊥ BC

AB² = AD² + BD²

AC² = AD² + CD²

AB² + AC² = AD² + BD² + AD² + CD²

=> AB² + AC² = BD² + CD² + 2AD²

AD² = BD × CD  given

=> AB² + AC² = BD² + CD² + 2BD × CD

=> AB² + AC² = (BD + CD)²

QED