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Difficulty: HardTriangle Properties and Angle Theorems

In △ABC\triangle ABC, the measure of ∠B\angle B is 80∘80^\circ and the measure of ∠C\angle C is 40∘40^\circ. A point DD lies on side BCBC such that ADAD bisects ∠BAC\angle BAC, and a point EE lies on side ACAC such that AD=AEAD = AE. What is the measure, in degrees, of ∠CDE\angle CDE?

Answer: 35 degrees

Answer

35
The correct answer is 3535. By first finding that ∠BAC=60∘\angle BAC = 60^\circ, we use the angle bisector ADAD to find ∠CAD=30∘\angle CAD = 30^\circ. In △ADC\triangle ADC, we find the interior angle ∠ADC=110∘\angle ADC = 110^\circ. In the isosceles triangle △ADE\triangle ADE with AD=AEAD=AE, the base angles are ∠ADE=∠AED=75∘\angle ADE = \angle AED = 75^\circ. Finally, subtracting ∠ADE\angle ADE from ∠ADC\angle ADC gives ∠CDE=35∘\angle CDE = 35^\circ.

Step-by-Step Solution

1
Find the measure of the third angle of the main triangle, ∠BAC\angle BAC.
∠BAC=60∘\angle BAC = 60^\circ
The sum of the interior angles of any triangle is 180∘180^\circ. Therefore, ∠BAC=180∘−∠B−∠C=180∘−80∘−40∘=60∘\angle BAC = 180^\circ - \angle B - \angle C = 180^\circ - 80^\circ - 40^\circ = 60^\circ.
2
Determine the measure of the bisected angle ∠CAD\angle CAD.
∠CAD=30∘\angle CAD = 30^\circ
Since ADAD bisects ∠BAC\angle BAC, it divides the angle into two equal parts: ∠BAD=∠CAD=60∘2=30∘\angle BAD = \angle CAD = \frac{60^\circ}{2} = 30^\circ.
3
Calculate the interior angle ∠ADC\angle ADC in △ADC\triangle ADC.
∠ADC=110∘\angle ADC = 110^\circ
In △ADC\triangle ADC, the sum of angles is 180∘180^\circ. Therefore, ∠ADC=180∘−∠CAD−∠C=180∘−30∘−40∘=110∘\angle ADC = 180^\circ - \angle CAD - \angle C = 180^\circ - 30^\circ - 40^\circ = 110^\circ.
4
Find the base angles of the isosceles triangle ADEADE.
∠ADE=75∘\angle ADE = 75^\circ
Since AD=AEAD = AE, △ADE\triangle ADE is an isosceles triangle with vertex angle ∠DAE=30∘\angle DAE = 30^\circ. The two base angles, ∠ADE\angle ADE and ∠AED\angle AED, are equal. Thus, ∠ADE=180∘−30∘2=75∘\angle ADE = \frac{180^\circ - 30^\circ}{2} = 75^\circ.
5
Determine the final angle ∠CDE\angle CDE by subtraction.
∠CDE=35∘\angle CDE = 35^\circ
Since point EE lies on side ACAC, ray DEDE lies between rays DADA and DCDC. Therefore, ∠ADC=∠ADE+∠CDE\angle ADC = \angle ADE + \angle CDE. Rearranging gives ∠CDE=∠ADC−∠ADE=110∘−75∘=35∘\angle CDE = \angle ADC - \angle ADE = 110^\circ - 75^\circ = 35^\circ.

Key Concept

Applying triangle angle sum theorem, angle bisector properties, and isosceles triangle base angle properties to perform multi-step angle tracing.

Alternative Method

Use the exterior angle theorem on △ADC\triangle ADC at vertex DD: ∠ADB=∠CAD+∠C=30∘+40∘=70∘\angle ADB = \angle CAD + \angle C = 30^\circ + 40^\circ = 70^\circ. Then, since EE is on ACAC, AA, EE, and CC are collinear. In △ADE\triangle ADE, the exterior angle at EE is ∠DEC=∠DAE+∠ADE=30∘+75∘=105∘\angle DEC = \angle DAE + \angle ADE = 30^\circ + 75^\circ = 105^\circ. In △DEC\triangle DEC, the sum of angles is 180∘180^\circ, so ∠CDE=180∘−105∘−40∘=35∘\angle CDE = 180^\circ - 105^\circ - 40^\circ = 35^\circ.
Estimated Time:2m 30s
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Topics

Triangle Properties and Angle Theorems

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