Question

Pinku was a hard working student who used to learn without understanding. He was asked to construct a triangle say ABC and was given the base length of the triangle BC, one of the base angles say $\angle$ B and the sum of the other two sides (AB + AC). He went about constructing the triangle in the following way:

He drew the base BC with the given dimension, drew the $\angle$ B along the ray BX with the angle known to him already. He then took B as centre and (AB + AC) as radius and cuts an arc on the ray BX intersecting the ray at D.

He then joins D to C. He then draws a perpendicular bisector of the line DC and the perpendicular bisector intersecting on the ray intersects the ray at point A. The teacher then asked him as to why he did what he did, she started from the back and asked him as to how the intersection of the ray and the perpendicular bisector gives A.

Which of the following is the reason for him drawing the perpendicular bisector and intersecting it with the ray?

• A. Since the perpendicular bisector would get AD = AC which is like flipping the point D about the perpendicular bisector and merging it with point C
• B. To get point A is the mid point of the line segment BD
• C. To get ∆ ABC and ∆ ADC congruent
• D. None of the above

Answer 1

The correct option is A

Since the perpendicular bisector would get AD = AC which is like flipping the point D about the perpendicular bisector and merging it with point C

Pinku drew the perpendicular bisector so as to have AD = AC, Pinku knew that BD = AB + AC and if he got AC = AD he'd have the triangle ABC.
For this to happen, he took the perpendicular bisector of line DC and intersected it at the ray BX and named the point of intersection as A.
If we take triangles AMC and AMD,

DM=MC (perpendicular bisector bisects the line DC)

$\angle$ AMC = $\angle$ AMD = ${90}^{\circ }$ (perpendicular bisector)

AM = AM (common side)

$△$AMC $\cong$ $△$AMD (By SAS Congruency)

Which means AC = AD (By CPCT)

Since AC = AD and BD = AB + AC, we have found the point A correctly so as to have the triangle ABC.