Question

1. segment ge is an angle bisector of both angle hef and angle fgh. prove triangle hge is congruent to triangle fge. 4. triangles acd and bcd are isosceles. angle bac has a measure of 33 degrees and angle bdc has a measure of 35 degrees. find the measure of angle abd. (from unit 2, lesson 6.) 5. which conjecture is possible to prove? a. all triangles with at least one side length of 5 are congruent. b. all pentagons with at least one side length of 5 are congruent. c. all rectangles with at least one side length of 5 are congruent. d. all squares with at least one side length of 5 are congruent. (from unit 2, lesson 5.)

Explanation:

3. Prove \(\triangle HGE\cong\triangle FGE\)

Step1: Define angle - bisector

Since \(GE\) is the angle bisector of \(\angle HEF\), we have \(\angle HEG=\angle FEG\). Since \(GE\) is the angle bisector of \(\angle FGH\), we have \(\angle HGE = \angle FGE\).

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

In \(\triangle HGE\) and \(\triangle FGE\), we have:

• \(GE = GE\) (common side)
• \(\angle HEG=\angle FEG\) (from the angle - bisector property)
• \(\angle HGE=\angle FGE\) (from the angle - bisector property)

By the \(ASA\) congruence criterion (\(ASA\): if two angles and the included side of one triangle are equal to two angles and the included side of another triangle, then the two triangles are congruent), \(\triangle HGE\cong\triangle FGE\).

4. Find the measure of \(\angle ABD\)

Step1: Analyze \(\triangle ACD\)

Since \(\triangle ACD\) is isosceles (\(AD = CD\)), \(\angle CAD=\angle ACD = 33^{\circ}\). Then, using the angle - sum property of a triangle (\(\angle CAD+\angle ACD+\angle ADC=180^{\circ}\)), we find \(\angle ADC=180-(33 + 33)=114^{\circ}\).

Step2: Analyze \(\angle BDC\) and \(\angle ADB\)

We know \(\angle BDC = 35^{\circ}\), so \(\angle ADB=\angle ADC-\angle BDC=114 - 35=79^{\circ}\).

Step3: Analyze \(\triangle BCD\)

Since \(\triangle BCD\) is isosceles (\(BD = CD\)), \(\angle BCD=\angle CBD\). Using the angle - sum property of a triangle (\(\angle BDC+\angle BCD+\angle CBD = 180^{\circ}\)), and since \(\angle BCD=\angle CBD\), we have \(\angle CBD=\frac{180 - 35}{2}=72.5^{\circ}\).

Step4: Analyze \(\triangle ABC\)

Since \(\triangle ACD\) is isosceles (\(AD = CD\)) and \(\triangle BCD\) is isosceles (\(BD = CD\)), \(AD = BD\). Let \(\angle ABD=x\). In \(\triangle ABD\), using the angle - sum property (\(\angle ABD+\angle BAD+\angle ADB=180^{\circ}\)). \(\angle BAD = 33^{\circ}\), \(\angle ADB = 79^{\circ}\). So \(x+33 + 79=180\). Then \(x=180-(33 + 79)=68^{\circ}\).

5. Determine which conjecture is provable

• For option A: Triangles with at least one side length of \(5\) are not congruent. Congruence of triangles requires \(SSS\), \(SAS\), \(ASA\), \(AAS\) or \(HL\) (for right - angled triangles). Just one side length is not sufficient. For example, a triangle with sides \(5,6,7\) and a triangle with sides \(5,8,9\) are not congruent.
• For option B: Pentagons with at least one side length of \(5\) are not congruent. Congruence of polygons requires that all corresponding sides and angles are equal. One side length is not enough. Different pentagons with one side of length \(5\) can have different other side lengths and angles.
• For option C: Rectangles with at least one side length of \(5\) are not congruent. Rectangles have opposite sides equal. If one side is \(5\), the adjacent side can have different lengths. For example, a rectangle with sides \(5,6\) and a rectangle with sides \(5,7\) are not congruent.
• For option D: Squares have all sides equal. If a square has at least one side length of \(5\), then all its sides are \(5\) (by the definition of a square: \(a = b = c = d\), where \(a,b,c,d\) are side lengths). And all angles of a square are \(90^{\circ}\). By \(SSS\) (since all four sides are equal and all four angles are equal), all squares with side length \(5\) are congruent.

Answer:

1. \(\triangle HGE\cong\triangle FGE\) by \(ASA\) criterion.
2. The measure of \(\angle ABD\) is \(68^{\circ}\).
3. D. All squares with at least one side length of \(5\) are congruent.