Question

2. ABC is a right triangle right angled at B.
Let D and E be any points on AB and
BC respectively.
(AS)
Prove that AE? + CD= AC? + DE.​

Answer 1

Step-by-step explanation:

As ΔABE is right angled triangle, then,

⇒ (AE)²=(AB)²+(BE)² ............(1)

Similarly, ΔDBC is right angled triangle, so,

⇒ (CD)²=(BD)²+(BC)² ............(2)

  Add equation (1) and (2),

⇒ (AE)²+(CD)²=(AB²+BE²)+(BD²+BC²)

⇒ (AE)²+(CD)²=(AB²+BC²)+(BD²+BE²) ............(3)

In ΔABC,

AC²=AB²+BC²

In ΔDBE,

⇒ DE²=BE²+BD²

So, equation (3) becomes,

⇒ (AE)²+(CD)²=(AC)²+(DE)²

    [Hence proved]