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Question
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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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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