Question

pls this questions. ​

Answer 1

A triangle ABC in which AB = AC and D is any point in BC.
To Prove:
AB2 - AD2 = BD.CD

Const: Draw AE ⊥ BC
Proof : In ∆ABE and ∆ACE, we have
AB = AC    [given]
AE = AE    [common]
and    ∠AEB = ∠AEC    [90°]
Therefore, by using RH congruent condition
∆ABE ~ ∆ACE
⇒    BE = CE
In right triangle ABE.
AB2 = AE2 + BE2 ...(i)
[Using Pythagoras theorem]
In right triangle ADE,
AD2 = AE2 + DE2
[Using Pythagoras theorem]
Subtracting (ii) from (i), we get
AB2 - AD2 = (AE2 + BE2) - (AE2 + DE2)
 AB2 - AD2 = AE2 + BE2 - AE2 - DE2
⇒ AB2 - AD2 = BE2 - DE2
⇒ AB2 - AD2 (BE + DE) (BE - DE)
But    BE = CE    [Proved above]
⇒ AB2 - AD2 = (CE + DE) (BE - DE)
= CD.BD
⇒ AB2 - AD2 = BD.CD Hence Proved.

HOPE IT WILL HELP UU MATE....... ;-D