Question

AB=AC&D is any point on BC AE is perpendicular to BC then show that AB²=BD×CD​

Answer 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]

Step-by-step explanation:

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