Question

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 $\angle B$ and angle MYX equal to $\angle C$

3. Bisect $\angle LXY$ and $\angle MXY$. 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?

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

The correct option is C

2

If we take $△$ XBQ and $△$ ABQ, we have

XQ = AQ and $\angle$ BQX = $\angle$ AQB = 90° (Since DE is the perpendicular bisector of AX)

BQ = BQ (common side)

$△$ XBQ $\cong$ $△$ ABQ (By SAS congruency)

XB = AB (By C.P.C.T.)

Similarly, taking $△$ ACR and $△$ YCR, we have

AR = YR and $\angle$ ARC = $\angle$ YRC = 90° (Since FG is the perpendicular bisector AY)

CR = CR (common side)

$△$ ACR $\cong$ $△$ YCR (By SAS Congruency)

AC = CY (By C.P.C.T.)

Hence the triangle ABX and triangle ACY are isosceles triangles.

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