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

Question

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

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Lyla 10 months 2021-07-24T06:43:57+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-24T06:45:46+00:00

    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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