105
In the given figure, DEFG is a square. $\triangle ABC$ is right angle triangle with $\angle A = 90^\circ$. Prove that AG $\times$ DG = AF $\times$ DB.
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DEFG is a square. $\therefore$ GF || BE (I) (1/2 Mark)
$\angle GBD = \angle AGF$ (II) (1/2 Mark)
$\therefore \triangle AGF \sim \triangle BDG$ (III) (1/2 Mark)
$\Rightarrow \frac{AG}{DB} = \frac{AF}{DG}$ (IV) (1/2 Mark)
$\Rightarrow AG \times DG = AF \times DB$
$\angle GBD = \angle AGF$ (II) (1/2 Mark)
$\therefore \triangle AGF \sim \triangle BDG$ (III) (1/2 Mark)
$\Rightarrow \frac{AG}{DB} = \frac{AF}{DG}$ (IV) (1/2 Mark)
$\Rightarrow AG \times DG = AF \times DB$