Finding the hypotenuse of a triangle using angles and segment lengths.

Question

In the diagram, $\angle CAB = 90^\circ.$ Let $D$ be a point on $\overline{AB},$ and let $E$ be a point on $\overline{AC},$ such that $AB = AC = DE,$ $BD = 9,$ and $CE = 8.$ Find $DE.$

[asy] unitsize(3 cm);

pair A, B, C, D, E;

A = (0,0); B = dir(230); C = dir(230 + 90); D = 0.6*B; E = 0.8*C;

draw(B--A--C); draw(D--E); draw(rightanglemark(B,A,C,2));

label("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); label("$D$", D, NW); label("$E$", E, NE); [/asy]

enter image description here

Answer 1

Let $x=AB = AC = DE $ and apply the Pythagorean theorem ${AE}^2+{AD}^2={DE}^2$ to the triangle $ADE$, i.e.

$$(x-9)^2+(x-8)^2=x^2$$

which yields $DE=x=29$.

Answer 2

Let $AB=AC=DE=x$ then we have $DB=9 and EC=8$ which gives us $AD=x-9\quad and\quad AE=x-8$ and by pythagoras formula in $\triangle ADE $ we get that $${AE}^2+{AD}^2={DE}^2$$ which gives us $${(x-9)}^2+{(x-8)}^2={(x)}^2$$ and it simplifies to $$x^2-34x+145=0$$ which gives us $$x=5 \text{or} x=29$$ but as we know $AD\ge0$ $x=5$ is eliminated and we have $x=29$ which implies $$DE=29$$