[tex] \frac{1}{2}[/tex]will be the answer.
Step-by-step explanation:-
If the altitude to the hypotenuse of a right-angled triangle bisects the hypotenuse, then we can know that the triangle is Isosceles right triangle. (AB=AC)
Make AB=AC=2, so,
BC=[tex] \sqrt{ {2}^{2} + {2}^{2} } [/tex]= [tex] \sqrt{4 + 4} [/tex]= [tex] \sqrt{8} [/tex]
Prime factorizing, we get,
BC = [tex] \sqrt{2 \times 2 \times 2} [/tex]
Take two 2's outside by taking common because of "square" root,
⇢BC = [tex]2 \sqrt{2} [/tex]
BD = DC = [tex] \sqrt{2} [/tex],
AD = [tex] \sqrt{AB^{2} - BD^{2} } [/tex]= [tex] \sqrt{4 - 2} [/tex]= [tex] \sqrt{2} [/tex]
So, AD=BD=CD= [tex] \frac{1}{2}[/tex]BC
∴⇢[tex] \frac{AD}{BC} = \frac{1}{2} [/tex]
Read more about hypotenuse altitude question;
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