Question

if two isosceles triangles have a common base prove that the line segment joining their vertices bisects the the common base at right angle ​

Answer 1

Answer:

This is the perpendicular bisector theorem, which you will use again and again in geometry class: 

The set of points equidistant to A and B is a line which is perpendicular to AB and bisects AB. 

If you already can use that theorem, showing that it applies to the triangles is easy: 

Call the endpoints of the base A and B. Call the vertices C and D. 

By the definition of an isosceles triangle, AC = BC and AD = BD. 

Therefore point C is equidistant to A and B, and point D is equidistant to A and B. 

Therefore C and D lie on the perpendicular bisector of AB. Therefore the line joining them is that perpendicular bisector, which of course is perpendicular to and bisects AB. 

If you're doing this exercise as a precursor to learning the perpendicular bisector theorem, then the proof is a bit slower. 

Again, call the endpoints of the base A and B and call the vertices C and D. Call the point of intersection E. 

AC = BC and AD = BD (as above). 

CD = CD. 

Therefore triangles ACD and BCD are congruent (all three side lengths are equal). 

Therefore angle ACD has the same measure as angle BCD. 

If you've drawn this you can see that angle ACD is another name for angle ACE, and BCD is BCE. 

Angle AEC = 180 - angle EAC - angle ACE. 

Angle BEC = 180 - angle EBC - angle BCE. 

We just proved that ACE = BCE. 

Because the triangle is isosceles, EAC = EBC. 

Therefore AEC = BEC. 

But angles AEC and BEC are supplementary. Since they are equal, they must both be 90 degrees. 

With all three angles and two sides proven equal, triangles ACE and BCE must be congruent. 

Therefore the final sides are also equal: AE = BE.

Hope this helps!