EXAMPLE 44 In Fig. 4.127, if $AD\perp BC and \frac{BD}{DA}=\frac{DA}{DC}$, prove that $△$ABC is a right triangle.

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Please find below the solution to the asked query:

We form our diagram , As :

Here given AD $\perp$ BC  , So 

$\angle$ ADB  =  $\angle$ ADC  =  90$°$                                               --- ( 1 )

As given : $\frac{\mathrm{BD}}{\mathrm{DA}} = \frac{\mathrm{DA}}{\mathrm{DC}} , \mathrm{So} ∆ \mathrm{BDA} ~ ∆ \mathrm{ADC}$ , Then

$\angle$ DBA = $\angle$ DAC                                                            --- ( 2 )       ( By C.P.S.T. )

And

$\angle$ DAB = $\angle$ DCA                                                            --- ( 3 )       ( By C.P.S.T. )

From angle sum property of triangle we get in triangle ADB :

$\angle$ DAB + $\angle$ DBA + $\angle$ ADB = 180$°$  , Substitute value from equation 1 we get :

$\angle$ DAB + $\angle$ DBA + 90$°$ = 180$°$  ,

$\angle$ DAB + $\angle$ DBA = 90$°$     , Substitute value from equation 3 we get :

$\angle$ DCA + $\angle$ DBA = 90$°$     ,

$\angle$ ACB + $\angle$ ABC = 90$°$                                                   --- ( 4 )      ( We know :$\angle$ DCA = $\angle$ ACB and  $\angle$ DBA =$\angle$ ABC same angles )


From angle sum property of triangle we get in triangle ABC :

$\angle$ BAC + $\angle$ ACB + $\angle$ ABC = 180$°$  , Substitute value from equation 4 we get :

$\angle$ BAC + 90$°$ = 180$°$ 

$\angle$ BAC = 90$°$  , So

Triangle ABC is a right angle triangle .                                        ( Hence proved )




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