Question

prove AJ is congruent to CJ​

Answer 1

The centre of the given circle be O.

The tangent at D meets AC at J . Join AD.

We have to prove that AJ = CJ.

$Since, \angle {ADJ}\: and \\ \angle {ABD} = \angle {ABC} \: are \: angles \: in \:the \: alternate \\segments \: of \: chord \: AD.$

$\therefore \angle {ADJ} = \angle {ABC} \: ---(1)$

$Since, AB \: is \: a \: diameter \: of \: the \\ circle\:and \: angle \:in \: a \: semicircle \: is \\right\:angle .$

$\therefore \angle {BDA} = 90\degree$

$\angle {ADC} = 90\degree \: ---(2)$

$( \ angle {BDA} \:and \: \angle {BDA} \:are \: linear \:pair )$

Since, ∆ABC is a right triangle , Right angle at A.

$\angle {ABC} + \angle {ACB} = 90\degree \: ---(3)$

/* From equations (2) and (3) , we get */

$\angle {ABC} + \angle {ACB} = \angle {ADC}$

$\implies \angle {ABC} + \angle {ACB} = \angle {ADJ} + \angle {JDC}$

$[ Since , \angle {ADC} = \angle {ADJ} + \angle {JDC} ]$

$\implies \angle {ABC} + \angle {ACB} = \angle {ABC} + \angle {JDC}$

$( Since, \angle {ADJ} = \angle {ABC} )$

$\implies \angle {ACB} = \angle {JDC}$

$\implies \angle {JCB} = \angle {JDC}$

$( Since , \angle {ACB} = \angle {JCB} )$

In ∆DJC , we have

$\implies \angle {JCB} = \angle {JDC}$

$\implies AJ = JC$

$\blue { Sides \: opposite \: to \: equal \: angles}\: ---(4)$

$\blue { are \: equal }$

Since, tangents from an exterior point to a circle are equal in length.

Therefore, JD = JA i.e DJ = JA ----(5)

From equations ( 4 ) and ( 5 ) , we get

AJ = JC .

Thus, J is a point on AC such that AJ = JC .

Hence , AJ congruent to JC.

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