QUESTION IMAGE
Question
recall that the sum of all of the angles of a triangle is 180° regardless of the size or type of triangle. use that fact and facts about isosceles triangles to answer the following.
- in △abc shown below, ac = bc.
(a) which two angles must be equal in this isosceles triangle?
(b) if m∠a = 56° then find m∠c.
- in the diagram below, rectangle abcd has point e located on side (overline{cd}) such that ae = be. if m∠dae = 61°, then which of the following is m∠aeb?
(1) 29° (2) 61° (3) 119° (4) 122°
- the ratio of the vertex angle to either of the base angles in an isosceles triangle is 5:2. determine the measure of the vertex angle. show how you arrived at your answer.
reasoning
- if we are told that (overline{cd}) is the perpendicular bisector of (overline{ab}) which of the following can we not conclude based on this information? explain your choice.
(1) ac = bc (2) ac = ad (3) m∠aec = m∠bec (4) ae = eb.
explanation:
Step1: Recall isosceles - triangle property
In an isosceles triangle, if two sides are equal, the angles opposite those sides are equal. In $\triangle ABC$ with $AC = BC$, $\angle A=\angle B$.
Step2: Use angle - sum property of a triangle
The sum of the angles in a triangle is $180^{\circ}$. Given $\angle A = 56^{\circ}$, and since $\angle A=\angle B = 56^{\circ}$, then $\angle C=180^{\circ}-\angle A-\angle B=180^{\circ}-56^{\circ}-56^{\circ}=68^{\circ}$.
Step3: Analyze rectangle and isosceles - triangle in rectangle $ABCD$
In rectangle $ABCD$, $\angle D = 90^{\circ}$. Given $\angle DAE=61^{\circ}$, then $\angle AED = 180^{\circ}-90^{\circ}-61^{\circ}=29^{\circ}$. Since $AE = BE$, $\triangle AEB$ is isosceles. $\angle AEB=180^{\circ}-2\times29^{\circ}=122^{\circ}$.
Step4: Solve for angles in isosceles triangle using ratio
Let the vertex angle be $5x$ and each base angle be $2x$. Then $5x + 2x+2x=180^{\circ}$ (angle - sum property of a triangle). Combining like terms gives $9x = 180^{\circ}$, so $x = 20^{\circ}$. The vertex angle is $5x=100^{\circ}$.
Step5: Analyze properties of perpendicular bisector
If $CD$ is the perpendicular bisector of $AB$, then any point on $CD$ is equidistant from $A$ and $B$. So $AC = BC$, $AE=EB$, and $\angle AEC=\angle BEC = 90^{\circ}$. But we cannot conclude that $AC = AD$.
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(a) $\angle A$ and $\angle B$
(b) $68^{\circ}$
- (4) $122^{\circ}$
- $100^{\circ}$
- (2) $AC = AD$. Explanation: A point on the perpendicular bisector of a line - segment is equidistant from the endpoints of the line - segment, so $AC = BC$, $AE = EB$, and $\angle AEC=\angle BEC$, but there is no reason to conclude $AC = AD$.