In figure, $D E F G$ is a square and $∠ B A C = 90$ . Show that $D E^{2} = B D \times E C$

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Solution

$s i n c e D E F G i s s q u a r e t h u s D E = E F = F G = D G \to (1)$

$I n △ D B G \sim△ A B C$

$\frac{E C}{A C} = \frac{E F}{A B}$

$h e n c e E C = \frac{(E F) (A C)}{A B} \to (2)$

$I n △ E F C \sim△ A B C$

$\frac{B D}{B A} = \frac{G D}{C A}$

$h e n c e B D = \frac{(G D) (B A)}{C A} \to (3)$

$E C \times B D = \frac{(E F) (A C) (G D) (B A)}{(A B) (C A)} \to (f r o m (1) a n d (2))$

$E C \times B D = D E^{2} \to (f r o m (1))$

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In figure, DEFG is a square and ∠BAC=90. Show that DE2=BD×EC

since DEFG is square thus DE=EF=FG=DG→(1)In△DBG∼△ABCECAC=EFABhence EC=(EF)(AC)AB→(2)In△EFC∼△ABCBDBA=GDCAhence BD=(GD)(BA)CA→(3)EC×BD=(EF)(AC)(GD)(BA)(AB)(CA)→(from(1) and (2))EC×BD=DE2→(from(1))

In figure, $D E F G$ is a square and $∠ B A C = 90$ . Show that $D E^{2} = B D \times E C$