Given: PQ is a perpendicular bisector of side AB of the triangle ABC. Prove: Q is equidistant from A and B. - Noon Academy

Given: PQ is a perpendicular bisector of side AB of the triangle ABC. Prove: Q is equidistant from A and B.

Class 10 | Exercise 16A | ICSE | Loci (Locus and its Constructions) | Maths | Selina

Given: PQ is a perpendicular bisector of side AB of the triangle ABC. Prove: Q is equidistant from A and B.

Selina Solutions Concise Class 10 Maths Chapter 16 ex. 16(A) - 1

solution:

Selina Solutions Concise Class 10 Maths Chapter 16 ex. 16(A) - 2

Construction: Join AQ

Proof:

In

$\vartriangle AQP\text{ }and\text{ }\vartriangle BQP,$

$AP\text{ }=\text{ }BP\text{ }\left[ Given \right]$

$\angle QPA\text{ }=\text{ }\angle QPB\text{ }[Each\text{ }{{90}^{o}}]$

$PQ\text{ }=\text{ }PQ\text{ }\left[ Common \right]$

So, by SAS criterion of congruence

$\vartriangle AQP\text{ }\cong \text{ }\vartriangle BQP$

By Corresponding Parts of Congruent Triangle (CPCT)

$AQ\text{ }=\text{ }BQ$

So,

$Q\text{ }is\text{ }equidistant\text{ }from\text{ }A\text{ }and\text{ }B.$

i

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