In triangle АВС, АВ = Ac
– D – С. ses AF I seg BC
4
Prove:
АB^2-АD^2 =
BD*CD​

1 thought on “In triangle АВС, АВ = Ac<br />– D – С. ses AF I seg BC<br />4<br />Prove:<br />АB^2-АD^2 =<br />BD*CD​”

1. Answer:

   Draw AE⊥BC

   In △AEB and △AEC, we have

   AB=AC

   AE=AE [common]

   and, ∠b=∠c [because AB=AC]

   ∴ △AEB≅△AEC

   ⇒ BE=CE

   Since △AED and △ABE are right-angled triangles at E.

   Therefore,

   AD

   2

   =AE

   2

   +DE

   2

   and AB

   2

   =AE

   2

   +BE

   2

   ⇒ AB

   2

   −AD

   2

   =BE

   2

   −DE

   2

   ⇒ AB

   2

   −AD

   2

   =(BE+DE)(BE−DE)

   ⇒ AB

   2

   −AD

   2

   =(CE+DE)(BE−DE) [∵BE=CE]

   ⇒ AB

   2

   −AD

   2

   =CD.BD

   AB

   2

   −AD

   2

   =BD.CD [Hence proved]

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