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If bisects the angle ACD then B is the midpoint of AD?
That is not necessarily true.
If A and B are two points in the plane the perpendicular bisector of line segment AB is the set of all points equidistant from A and B?
Let us consider a line l such that it is the perpendicular bisector of line segment AB and the line intersects at point C.Let us take any point on line l(say, D). Join A to D and B to D.Now we have two triangles ACD and BCD.Now, in triangles ACD & BCD, we haveCD = CD (Common side)�ACD = �BCD (Right angle)AC = BC (Since l bisects AB)According to Side-Angle-Side criteria: Both triangles are congruent.Since both triangles are congruent, therefore AD = BD.So, l is the set of all points equidistant from A & B.Hence proved.
Angle bisector of angleA of triangleABC is perpendicular to BC prove it is isosceles?
Let D represent the point on BC where the bisector of A intersects BC. Because AD bisects angle A, angle BAD is congruent to CAD. Because AD is perpendicular to BC, angle ADB is congruent to ADC (both are right angles). The line segment is congruent to itself. By angle-side-angle (ASA), we know that triangle ADB is congruent to triangle ADC. Therefore line segment AB is congruent to AC, so triangle ABC is isosceles.
If the bisector of the vertical angle of the triangle bisects the base prove that the triangle is isosceles?
show that the bisector of the vertical angle of an isosceles triangle bisects the base
If segment AD perpendicular segment CE m angle 1 m angle 6 and m angle 1 37 find m angle GMD?
143
