Question

in triangle ABC,if AB=AC and D is mid point on BC. prove that AB2-AD2=BD X CD​

Answer 1

Answer:

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.