Question

1. given: \\( \overline { l t } \cong \overline { t r } \\), \\( \angle l t \cong \angle e t r \\), \\( i t \parallel e r \\) prove: \\( \triangle l i t \cong \triangle t e r \\) 4. given: \\( c \\) is midpoint of \\( \overline { b d } \\) \\( \overline { a b } \perp \overline { b d } \\) \\( \overline { b d } \perp \overline { d e } \\) prove: \\( \triangle a b c \cong \triangle e d c \\) 5. given: point \\( p \\) is on the perpendicular bisector of segment \\( \overline { a b } \\) prove: \\( p \\) is equidistant from points \\( a \\) and \\( b \\)

Explanation:

3. Prove: $\triangle LIT\cong\triangle TER$

Step1: Find alternate - interior angles

Since \(IT\parallel ER\), by the Alternate - Interior Angles Theorem, \(\angle ITL=\angle ERT\).

Step2: Use ASA (Angle - Side - Angle) congruence criterion

We are given that \(\overline{LT}=\overline{TR}\), \(\angle ILT = \angle ETR\), and we just found that \(\angle ITL=\angle ERT\).
In \(\triangle LIT\) and \(\triangle TER\):

• \(\angle ILT=\angle ETR\) (Given)
• \(\overline{LT}=\overline{TR}\) (Given)
• \(\angle ITL=\angle ERT\) (Alternate - Interior Angles Theorem)

By the ASA congruence criterion, \(\triangle LIT\cong\triangle TER\).

4. Prove: \(\triangle ABC\cong\triangle EDC\)

Step1: Use the definition of mid - point

Since \(C\) is the mid - point of \(\overline{BD}\), then \(\overline{BC}=\overline{DC}\) (Definition of mid - point: A mid - point of a segment divides the segment into two equal parts).

Step2: Use the definition of perpendicular lines

Since \(\overline{AB}\perp\overline{BD}\) and \(\overline{BD}\perp\overline{DE}\), then \(\angle ABC=\angle EDC = 90^{\circ}\) (Definition of perpendicular lines: If two lines are perpendicular, the angle between them is \(90^{\circ}\)).

Step3: Use the Vertical Angles Theorem

\(\angle ACB=\angle ECD\) (Vertical Angles Theorem: Vertical angles are congruent).

Step4: Use ASA congruence criterion

In \(\triangle ABC\) and \(\triangle EDC\):

• \(\angle ABC=\angle EDC\) (From Step 2)
• \(\overline{BC}=\overline{DC}\) (From Step 1)
• \(\angle ACB=\angle ECD\) (From Step 3)

By the ASA congruence criterion, \(\triangle ABC\cong\triangle EDC\).

5. Prove: \(P\) is equidistant from Points \(A\) and \(B\)

Step1: Let \(L\) be the perpendicular bisector of \(\overline{AB}\) and \(M\) be the mid - point of \(\overline{AB}\)

Since \(P\) is on the perpendicular bisector of \(\overline{AB}\), \(\overline{PM}\perp\overline{AB}\) and \(\overline{AM}=\overline{BM}\) (Definition of perpendicular bisector: A perpendicular bisector of a segment is perpendicular to the segment and passes through its mid - point).

Step2: Consider right triangles \(\triangle PAM\) and \(\triangle PBM\)

In \(\triangle PAM\) and \(\triangle PBM\):

• \(\overline{AM}=\overline{BM}\) (From Step 1)
• \(\overline{PM}=\overline{PM}\) (Common side)
• \(\angle PMA=\angle PMB = 90^{\circ}\) (From Step 1)

Step3: Use HL (Hypotenuse - Leg) congruence criterion for right triangles

By the HL congruence criterion (For right triangles, if the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of another right triangle, the two right triangles are congruent), \(\triangle PAM\cong\triangle PBM\).

Step4: Use the property of congruent triangles

Since \(\triangle PAM\cong\triangle PBM\), then \(\overline{PA}=\overline{PB}\) (Corresponding parts of congruent triangles are congruent). So \(P\) is equidistant from \(A\) and \(B\).

Answer:

1. \(\triangle LIT\cong\triangle TER\) by ASA.
2. \(\triangle ABC\cong\triangle EDC\) by ASA.
3. \(P\) is equidistant from \(A\) and \(B\) (i.e., \(\overline{PA}=\overline{PB}\)) as shown above.