In triangle АВС, АВ = Ac
- D - С. ses AF I seg BC
4
Prove:
АB^2-АD^2 =
BD*CD
Answers
Answered by
9
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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