A student constructed a triangle with the known conditions to him being the perimeter of the triangle and both the base angles. The steps of construction he used are as follows:
1. Draw a line segment, say XY equal to AB + BC + AC
2. Make angles LXY equal to ∠B and angle MYX equal to ∠C
3. Bisect ∠LXY and ∠MXY. Let these bisectors intersect each other at A.
4. Draw perpendicular bisectors DE of AX and FG of AY
5. Let DE intersect XY at B and FG intersect XY at C. Join AB and AC.
6. The triangle ABC is thus formed.
How many isosceles triangles are there in this figure?
If we take △ XBQ and △ ABQ, we have
XQ = AQ and ∠ BQX = ∠ AQB = 90° (Since DE is the perpendicular bisector of AX)
BQ = BQ (common side)
△ XBQ ≅ △ ABQ (By SAS congruency)
XB = AB (By C.P.C.T.)
Similarly, taking △ ACR and △ YCR, we have
AR = YR and ∠ ARC = ∠ YRC = 90° (Since FG is the perpendicular bisector AY)
CR = CR (common side)
△ ACR ≅ △ YCR (By SAS Congruency)
AC = CY (By C.P.C.T.)
Hence the triangle ABX and triangle ACY are isosceles triangles.