Question

drawing △abc with ab = bc. then he constructs a line perpendicular to ac that passes through b and intersects ac at d.

1. complete leroys proof that (overleftrightarrow{bd}) is the perpendicular bisector of (overline{ac}).

1. since (overleftrightarrow{bd}) is perpendicular to (overline{ac}), are both right triangles. by the, ((ad)^{2}+(bd)^{2}=(ab)^{2}). similarly,

. because ab = bc, by substitution.
by the property of equality, ((ad)^{2}+(bd)^{2}=(bd)^{2}+(cd)^{2}). this means ((ad)^{2}=(cd)^{2}) by the property of equality. therefore, and (overleftrightarrow{bd}) is the perpendicular bisector of (overline{ac}).

Explanation:

Step1: Identify right - triangles

Since $\overrightarrow{BD}$ is perpendicular to $\overline{AC}$, $\triangle ABD$ and $\triangle CBD$ are both right triangles.

Step2: Apply Pythagorean theorem

By the Pythagorean theorem, $(AD)^{2}+(BD)^{2}=(AB)^{2}$. Similarly, $(CD)^{2}+(BD)^{2}=(BC)^{2}$.

Step3: Use the given equality

Because $AB = BC$, $(AB)^{2}=(BC)^{2}$ by substitution.

Step4: Substitute and simplify

By the substitution property of equality, $(AD)^{2}+(BD)^{2}=(BD)^{2}+(CD)^{2}$. This means $(AD)^{2}=(CD)^{2}$ by the subtraction property of equality.

Step5: Conclude the result

Therefore, $AD = CD$ and $\overrightarrow{BD}$ is the perpendicular bisector of $\overline{AC}$.

Answer:

$\triangle ABD$ and $\triangle CBD$; Pythagorean theorem; $(CD)^{2}+(BD)^{2}=(BC)^{2}$; $(AB)^{2}=(BC)^{2}$; substitution; subtraction; $AD = CD$