DE = AB, EF = BC and AC = DF, hence triangle ABC is congruent to triangle DEF.
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Congruent shape</h3>
Two shapes are said to be congruent if they have the same shape, all their corresponding angles and sides are congruent to one another.
Given that DE = AB and BC = EF.
In right triangle DEF, using Pythagoras:
DF² = DE² + EF²
Also, In right triangle ABC, using Pythagoras:
AC² = AB² + BC²
But DE = AB and EF = BC, hence:
AC² = DE² + EF²
AC² = DF²
Taking square root of both sides, hence:
AC = DF
Since DE = AB, EF = BC and AC = DF, hence triangle ABC is congruent to triangle DEF.
Find out more on Congruent shape at: brainly.com/question/11329400
3y+14=44
Answer: First step is to subtract 14 from each side
3y = 30 (divide each side by 3)
y = 30/3, y = 10
Step-by-step explanation:
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The tangent line is perpendicular to the radius from the center of the circle to the point of tangency.
ΔDCE is similar to ΔDBA
DC/DB=DE/DA
15/50=DE/40
DE=12
