In triangle ABC, angle A is 30 degrees, angle B is 88 degrees, CD is the bisector of the outer angle at vertex C, and point D lies on line AB. On the continuation of the AC side beyond point C, a point E is chosen such that CE = CB. Find the corner BDE.
In a triangle, the degree measure of the outside angle is equal to the sum of the other two inside angles that are not adjacent to it. We get:
Angle ВСЕ = 30 ° + 88 ° = 118 °.
Consider two triangles, BCD and ECD, which we have:
ВС = EU (by condition);
CD – common side;
Angle ВСD = angle ECD = 118 ° / 2 ° = 59 ° (CD – bisector).
From all this we conclude that the triangles are equal.
In the triangle BCD we find the angle B, it is adjacent to the inner angles of the triangle ABC:
Angle B = 180 ° – 88 ° = 92 °.
In the same triangle, we find the angle BDC:
Angle BDC = 180 ° – (92 ° + 59 °) = 29 °.
Angle BDC = angle EDC (triangle equal), we get that;
angle BDE = 2 * 29 ° = 58 °.
Answer: The BDE is 58 °.
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