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 MYX equal to ∠ C
3. Bisect ∠ LXY and ∠ MXY. Let these bisectors intersect each other at A.
4. Draw perpendicular bisectors PQ of AX and RS of AY
5. Let PQ intersect XY at B and RS 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 (perpendicular bisector)
∠ BQX = ∠ AQB = 90° (perpendicular bisector)
BQ = BQ (common side)
△ XBQ ≅ △ ABQ (By SAS Congruency)
XB = AB (By CPCT)
Similarly, taking △ ACR and △ YCR, we have
AR = YR (perpendicular bisector)
∠ ARC = ∠ YRC = 90°(perpendicular bisector)
CR = CR (common side)
△ ACR ≅ △ YCR (By SAS Congruency)
AC = CY (By CPCT)
Hence the triangles ABX and ACY are isosceles triangles.