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Question

ABD is a right triangle right-angled at A and ACBD. Show that
(i) AB2=BC.BD
(ii) AC2=BC.DC
(iii) AD2=BDCD
(iv) AB2AC2=BDDC

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Solution


(i) In ADB and CAB

DAB=ACB [ Each 90o ]

ADB=CBA [ Common angle ]

ADBCAB

ABCB=BDAB

AB2=BC×BD [ Hence proved ]

(ii) Let CAB=x --- ( 1 )

In CBA

CBA=180o90ox

CBA=90ox --- ( 2 )

Similarly in CAD,

CAD=90oCAD

CAD=90ox ---- ( 3 )

CDA=180o90o(90ox)

CDA=x --- ( 4 )

In CBA and CAD

CBA=CAD [ From ( 2 ) and ( 3 ) ]

CAB=CDA [ From ( 1 ) and ( 4 ) ]

CBACAD [ By AA similarity ]

ACDC=BCAC

AC2=BC×DC ---- [ Hence proved ]


(iii) In DCA and DAB

DCA=DAB [ Each 90o]

CDA=ADB [ Common angle ]

DCADAB

DCDA=DADB

AD2=BD×CD ---- [ Hence proved ]


(iv) AB2AC2=BC×BDBC×DC

AB2AC2=BDDC

931620_969520_ans_4fb21c1004054cf1a46d04b547ccaf5c.png

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