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Magazine Chapter 3: Problem 19
Given that \(\triangle A B C\) is isosceles with \(\overline{A B} \approx \overline{A C}\), give the reason why \(\triangle A B D \cong \triangle A C D\) if \(D\) lies on \(\overline{B C}\) and: (a) \(\overline{A D}\) is an altitude of \(\triangle A B C\) (b) \(\overline{A D}\) is a median of \(\triangle A B C\). (c) \(\overline{A D}\) is the bisector of \(\angle B A C\)
Short Answer
Expert verified
Congruence in each part is shown using HL, SSS, and ASA theorems.
Step by step solution
01
Identify Given Facts
Since \( \triangle ABC \) is isosceles with \( \overline{AB} \cong \overline{AC} \), sides AB and AC are equal.
02
Analyze Condition (a): Altitude
If \( \overline{AD} \) is an altitude of \( \triangle ABC \), then \( AD \perp BC \) and \( \angle ADB = \angle ADC = 90^{\circ} \). Thus, \( \triangle ABD \) and \( \triangle ACD \) are right triangles with \( \overline{AD} \) being a common side.
03
Apply Hypotenuse-Leg Congruence
Since \( \overline{AB} = \overline{AC} \) and \( \overline{AD} \) is common between \( \triangle ABD \) and \( \triangle ACD \), the triangles are congruent by the Hypotenuse-Leg (HL) Congruence Theorem.
04
Analyze Condition (b): Median
If \( \overline{AD} \) is a median of \( \triangle ABC \), it implies that \( D \) is the midpoint of \( \overline{BC} \), so \( \overline{BD} \cong \overline{DC} \). \( \overline{AD} \) is a common side, and \( \overline{AB} \cong \overline{AC} \) by given.
05
Apply Side-Side-Side Congruence
With \( \overline{AB} = \overline{AC} \), \( \overline{BD} = \overline{DC} \) and \( \overline{AD} \) common, \( \triangle ABD \cong \triangle ACD \) by the SSS Congruence Theorem.
06
Analyze Condition (c): Angle Bisector
If \( \overline{AD} \) is the angle bisector of \( \angle BAC \), then \( \angle BAD = \angle CAD \). Sides \( \overline{AB} = \overline{AC} \) by definition, and \( \overline{AD} \) is common.
07
Apply Angle-Side-Angle Congruence
Since \( \angle BAD = \angle CAD \), \( \overline{AD} \) is a common side, and \( \overline{AB} = \overline{AC} \), triangles \( \triangle ABD \cong \triangle ACD \) by the ASA Congruence Theorem.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Isosceles Triangle
An isosceles triangle is a triangle that has at least two sides of equal length. In the case of triangle \( \triangle ABC \), sides \( \overline{AB} \) and \( \overline{AC} \) are of equal lengths. This property leads to several conclusions about the angles and the symmetry of the triangle.
- Since \( \overline{AB} \cong \overline{AC} \), this implies that the angles opposite these sides are also equal; thus, \( \angle ABC = \angle ACB \).
- Isosceles triangles often have additional properties such as symmetrical altitudes, medians, and angle bisectors, all originating from the vertex angle down to the base.
HL Congruence Theorem
The Hypotenuse-Leg (HL) Congruence Theorem is specific to right triangles. It states that if the hypotenuse and one leg of one right triangle are equal to the hypotenuse and leg of another right triangle, the triangles are congruent. For the isosceles \( \triangle ABC \) where \( \overline{AD} \) is an altitude, triangle \( \triangle ABD \) and \( \triangle ACD \) are right triangles:
- The hypotenuses are \( \overline{AB} \) and \( \overline{AC} \), which are equal because \( \triangle ABC \) is isosceles.
- \( \overline{AD} \) is a common leg for both triangles.
SSS Congruence Theorem
The Side-Side-Side (SSS) Congruence Theorem states that if three sides of one triangle are equal to three sides of another triangle, the triangles are congruent.Considering that \( \triangle ABC \) is isosceles and \( \overline{AD} \) is a median:
- \( \overline{AB} = \overline{AC} \) due to the isosceles nature of \( \triangle ABC \).
- \( \overline{BD} \cong \overline{DC} \) because \( D \) is the midpoint of \( \overline{BC} \).
- \( \overline{AD} \) is a common side.
ASA Congruence Theorem
The Angle-Side-Angle (ASA) Congruence Theorem is another principle for proving triangle congruence. According to ASA, if two angles and the included side of one triangle are equal to two angles and the included side of another triangle, the triangles are congruent.In the situation where \( \overline{AD} \) is the angle bisector of \( \angle BAC \):
- \( \angle BAD = \angle CAD \) because \( \overline{AD} \) is the bisector.
- \( \overline{AB} = \overline{AC} \) as given by the isosceles property.
- \( \overline{AD} \) is the side included between angles \( \angle BAD \) and \( \angle CAD \).
Altitudes
An altitude in a triangle is a line segment drawn from a vertex to the opposite side, intersecting it at a right angle. For \( \triangle ABC \), the altitude \( \overline{AD} \) means:
- \( AD \perp BC \).
- This creates two right angles, \( \angle ADB \) and \( \angle ADC \), making \( \triangle ABD \) and \( \triangle ACD \) right triangles.
Medians
A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. When \( \overline{AD} \) is a median in \( \triangle ABC \):
- Point \( D \) is the midpoint of \( \overline{BC} \), which means \( \overline{BD} = \overline{DC} \).
- \( \overline{AD} \) bisects \( \overline{BC} \), ensuring balance within the triangle.
Angle Bisectors
An angle bisector in a triangle divides an angle into two equal parts. Considering \( \overline{AD} \) as an angle bisector in \( \triangle ABC \):
- \( \angle BAD \) equals \( \angle CAD \).
- This proportional division allows use of the ASA Congruence Theorem.
