Question

Pallav got following question in his exam:

Given the base angles, say $\angle$ B and $\angle$ C and BC + CA + AB, construct triangle ABC.

He followed the following steps for the construction.

1. Draw a line segment, say XY equal to BC + CA + AB.

2. Make $\angle$ LXY equal to $\angle$ B and $\angle$ MYX equal to $\angle$ C.

3. Bisect $\angle$ LXY and $\angle$ MYX. Let these bisectors intersect at a point 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

• A. True
• B. False

Answer 1

The correct option is A

True

Yes, he has followed correct procedure.

1. Draw a line segment, say XY equal to BC + CA + AB.

2. Make angles $\angle$ LXY equal to $\angle$B and $\angle$ MYX equal to $\angle$C.

3. Bisect $\angle$LXY and $\angle$MYX. Let these bisectors intersect at a point 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

Then ABC is the required triangle. For the justification of the construction, you observe that B lies on the perpendicular bisector PQ of AX.

Therefore, XB = AB and similarly, CY = AC.

This gives BC + CA + AB = BC + XB + CY = XY.

Again $\angle$BAX =$\angle$ AXB (As in $△$ AXB, AB = XB) and

$\angle$ABC = $\angle$BAX + $\angle$AXB = 2 $\angle$AXB = $\angle$LXY [Using exterior angles property]

Similarly, $\angle$ACB = $\angle$MYX as required.