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
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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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