Question

In triangle abc angle b is equal to 90°. If d and e are any pts on ab and ac . Prove that ae^2+cd^2=ac^2+de^2

Answer 1

Answer:

$\bold{AE^2+DC^2=AC^2+DE^2}$

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.

By pythagors theorem,

In right ∆ABC, $AC^2=AB^2+BC^2$...(1)

In right∆ABE, $AE^2=AB^2+BE^2$...(2)

In right ∆DBC, $DC^2=DB^2+BC^2$...(3)

In right ∆DBE, $DE^2=DB^2+BE^2$...(4)

Now, adding (2) and (3) we get

$AE^2+DC^2=(AB^2+BE^2)+(DB^2+BC^2)$

$\implies\:AE^2+DC^2=(AB^2+BC^2)+(DB^2+BE^2)$

$\implies\:AE^2+DC^2=AC^2+(DB^2+BE^2)$ (using (1))

$\implies\:\bold{AE^2+DC^2=AC^2+DE^2}$ (using (4))