In a triangle ABC AB = BC; AC = 10 cm. From point D, the middle of AB, a perpendicular DE is drawn to side AB

| January 28, 2021 | education

In a triangle ABC AB = BC; AC = 10 cm. From point D, the middle of AB, a perpendicular DE is drawn to side AB until it intersects with BC at point E, and point E is connected to A. Perimeter of triangle ABC = 40 cm. Find the perimeter of triangle AEC.

1) Find the side of the ABC trips:
P triug ABC = 2 * AB + AC;
AB = BC = (P triug ABC – AC) / 2;
AB = BC = (40 – 10) / 2 = 30/2 = 15cm.
2) Consider the ADE and BDE trains:
AD = DB by task condition;
DE = common side;
Angle BDE = Angle ADE = 90 degrees.
According to the first sign of equality of triangles (two sides and the angle between them of one triangle are equal to two sides and the angle between them of another triangle) triangles ADE = triangles BDE.
3) Find the perimeter of the AEC trips:
P treug AEC = AC + AE + EC;
AE = BE;
P treug AEC = AC + (BE + EC) = AC + BC = 10 + 15 = 25 cm.



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