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State the locus of a point in a rhombus ABCD, which is equidistant from AB and AD; from the vertices A and C.

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Question

State the locus of a point in a rhombus ABCD, which is equidistant

  1. from AB and AD;
  2. from the vertices A and C.
Sum
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Solution

 
Steps of construction:

i. In rhombus ABCD, draw angle bisector of ∠A which meets in C.

ii. Join BD, which intersects AC at O.

O is the required locus.

iii. From O, draw OL ⊥ AB and OM ⊥ AD

In ΔAOL and ΔAOM

∠OLA = ∠OMA = 90°

∠OAL = ∠OAM  ...(AC is bisector of angle A)

AO = OA  ...(Common)

By Angle-Angle – side criterion of congruence,

ΔAOL ≅ ΔAOM  ...(AAS Postulate)

The corresponding parts of the congruent triangles are congruent

`=>` OL = OM ...(C.P.C.T.)

Therefore, O is equidistant from AB and AD.

Diagonal AC and BD bisect each other at right angles at O.

Therefore, AO = OC

Hence, O is equidistant from A and C. 

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