Pallav got following question in his exam:
Given the base angles, say ∠ B and ∠ 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 ∠ LXY equal to ∠ B and ∠ MYX equal to ∠ C.
3. Bisect ∠ LXY and ∠ 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
True
Yes, he has followed correct procedure.
1. Draw a line segment, say XY equal to BC + CA + AB.
2. Make angles ∠ LXY equal to ∠B and ∠ MYX equal to ∠C.
3. Bisect ∠LXY and ∠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 ∠BAX =∠ AXB (As in △ AXB, AB = XB) and
∠ABC = ∠BAX + ∠AXB = 2 ∠AXB = ∠LXY [Using exterior angles property]
Similarly, ∠ACB = ∠MYX as required.