Triangle ABC and the triangle EAD of the figure below are congruent triangles.
∠ABC=∠ADE = 40 degrees.
The intersection of the line connecting C and D and the side AE is set to P.
Find the angle of ∠APC.
Answer
70 degrees
Solution
As shown in the figure below, the auxiliary line connecting C and E is drawn.
∠BAC=180 - 40 × 2 = 100 degrees.
∠EAC=100 - 40 = 60 degrees.
Since AE = AC, △ACE is an equilateral triangle.
Thus, ∠AEC = 60 degrees.
∠AED = ∠BAC = 100 degrees.
∠CED = 60 + 100 = 160 degrees.
Since △ECD is an isosceles triangle with DE = CE, ∠ECD = (180 - 160) / 2 = 10 degrees. Therefore, ∠APC =∠PEC +∠ECD = 60 + 10 = 70 degrees.
∠ABC=∠ADE = 40 degrees.
The intersection of the line connecting C and D and the side AE is set to P.
Find the angle of ∠APC.
Answer
70 degrees
Solution
As shown in the figure below, the auxiliary line connecting C and E is drawn.
∠BAC=180 - 40 × 2 = 100 degrees.
∠EAC=100 - 40 = 60 degrees.
Since AE = AC, △ACE is an equilateral triangle.
Thus, ∠AEC = 60 degrees.
∠AED = ∠BAC = 100 degrees.
∠CED = 60 + 100 = 160 degrees.
Since △ECD is an isosceles triangle with DE = CE, ∠ECD = (180 - 160) / 2 = 10 degrees. Therefore, ∠APC =∠PEC +∠ECD = 60 + 10 = 70 degrees.
