You arrive at an intersection with traffic lights

Ответ:

arod20061

15.04.2020

correct choice is b

step-by-step explanation:

in right triangle cde, $m\angle cde=90^{\circ}.$ let $m\angle ecd=x^{\circ},$ then $m\angle dec=(90-x)^{\circ}.$

consider right triangle dae. in this triangle

$m\angle dae=90^{\circ};$ $m\angle dea=(90-x)^{\circ};$ $m\angle eda=90^{\circ}-(90-x)^{\circ}=x^{\circ}.$

then $\triangle cde\sim \triangle dae$ by aaa theorem (or $\triangle ead\sim \triangle edc$ )

consider right triangle dac. in this triangle

$m\angle dac=90^{\circ};$ $m\angle acd=x^{\circ};$ $m\angle cda=90^{\circ}-x^{\circ}.$

then $\triangle cde\sim \triangle cad$ by aaa theorem.

by transitive property,

$\triangle ead\sim \triangle edc\sim \triangle dac$ (correct option is b).

note that neither δbac nor δabc is similar to triangle ead, because these triangles can have arbitrary angles acb and cab.

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