Look at the figure below: triangle abc has measure of angle abc equal to 90 degrees. e is a point on the hypotenuse ac such that the segment be is perpendicular to ac. d is a point outside the triangle on the right of e so that triangle edc has the measure of angle edc equal to 90 degrees. which triangle is similar to triangle aeb?

δaeb δabc δ bec

step-by-step explanation:

given:   ∠abc= , => ∠aeb =   , and ∠edc =  

let ∠bac = x , then by angle sum property in δabc, ∠bca = 90°-x

now, we can see, in δaeb, ∠bae = x (∵∠bac = x)

so, again by angle sum property in δaeb , ∠abe = 180°-x - 90° = 90°-x

in δabc & δaeb

∠abc=∠aeb (∵each 90°)

∠bac = ∠bae (∵ each equals to x)

∠acb = ∠abe = 90°-x

so, by aaa criteria of similarity

δaeb is similar to δabc

by using transitive property,

δaeb δabc δ bec

here, neither δedc nor δcde is similar to δaeb, beause ∠edc and ∠cde are arbitrary.

congruent segments

step-by-step explanation:

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.