Want this question answered?
The answer is "No Solution" because there is not enough information.
the midpoint of
27
Label the triangle ABC. Draw the bisector of angle A to meet BC at D. Then in triangles ABD and ACD, angle ABD = angle ACD (equiangular triangle) angle BAD = angle CAD (AD is angle bisector) so angle ADB = angle ACD (third angle of triangles). Also AD is common. So, by ASA, triangle ABD is congruent to triangle ACD and therefore AB = AC. By drawing the bisector of angle B, it can be shown that AB = BC. Therefore, AB = BC = AC ie the triangle is equilateral.
Yes. AB, AC, BC and EF.
14
If 2 segments have the same length they are known as 'congruent segments' IE: segment AB=segment AC (or AB=AC) then AB @ AC (or AB is congruent to AC)
segment ac
The answer is "No Solution" because there is not enough information.
Ab+bc=ac
Since B is located between A and C, you can just add the two lengths together, so AC = m + n.your segment looks like this:A----B----Cwhere AB=m, BC=n, and AC=m+n
Usually a line segment in a triangle is either assigned a letter, or is referred to by the letters at the end of the segment with a line overhead. For instance, if you have a triangle ABC, you'll have segments AB, AC, and BC. And there would be a line over each segment name.
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.
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.
It is easiest to draw it using two right angled triangles.Draw a line AB that is 2 units long. From B, draw BC which is perpendicular to AB and 2 units long. Join AC. From C, draw CD which is perpendicular to AC (clockwise if BC is clockwise from AB, or anticlockwise if BC is anticlockwise) and make CD 2 uinits long. Then AD is a line segment which is sqrt(12) units long.
Assuming A, B and C are consecutive points on a straight line, AB would be 49 mm
C is not on the line AB.