Math, asked by pravallinowduri2006, 4 days ago

5. ABD is a triangle right angled at A and AC perpendicular to BD
Show that (i) AB^2 = BC.BD
(ii) AC^2 =BC.DC
(iii) AD^2= BD.CD

Answers

Answered by VishalRai46
2

Step-by-step explanation:

(i) In △BCA and △BAD,

∠BCA=∠BAD ....Each 90°

∠B is common between the two triangles.

So, △BCA∼△BAD ...AA test of similarity ....(I)

Hence,

BC/AB = AC/AD = AB/BD ...C.S.S.T

And, ∠BAC=∠BDA ....C.A.S.T ....(II)

so,

BC/AB= AB/BD

∴AB^2=BC×BD

Hence proved.

(ii) In △BCA and △DCA,

∠BCA=∠DCA ....Each 90°

∠BAC=∠CDA ...From (II)

So, △BCA∼△ACD ...AA test of similarity ....(III)

Hence, BC/AC = AC/CD = AB/BD

so,

BC/AC = AC/CD

∴AC ^2 =BC×DC

Hence proved.

(iii) From (I) and (III), we get

△BAD∼△ACD

Hence,AB/AC = AD/CD = BD/AD

So, AD ^2 =BD×CD

Hence proved.

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