On the continuation of the median AD, the triangle ABC was marked and the segment DE = AD was postponed. Prove that: 1) triangle ABD = triangle ECD; 2) Triangle ACD = Triangle EBD
By the condition of the problem, we are given that the segments AD and DE are equal to each other.
Since AD is the median, it divides the AC side into two equal segments (by the median property).
The segments BC and AE intersect at point D. This means that if we supplement the drawing with the sides BE and EC, we will get a parallelogram ABEC, because the diagonals of the parallelogram intersect and they are halved by the point of intersection.
In a parallelogram, the internal opposite angles are equal in pairs.
The angles BDA and EDC are vertical, which means they are equal.
The other four corners of these triangles are crosswise and are also equal.
Respectively with triangles BDE and ADC.
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