Question

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Answer 1

Answer:

(i) Yes,

In ΔBOC and ΔDOC

BO = DO [In a rhombus diagonals bisect each other]

CO = CO Common

BC = CD [All sides of a rhombus are equal]

By using SSS Congruency, ΔBOC≅ΔDOC.

(ii) Yes,

∠BCO = ∠DCO, by corresponding parts of congruent triangles.

Answer 2

Answer:

1) to prove:

∆AOB=∆ BOC

proof:

angle AOB=90°.........(1)

angle AOB+angle BOC=180°....(angles in linear pair)

90°+angle BOC=180°

angle BOC=180°-90°

angle BOC=90°.....(2)

from 1 and 2

angle AOB=angle BOC....(3)

in triangle AOB and BOC

seg AO=seg OC...(Given)

seg OB=seg OB....(Common side)

angle AOB=angle BOC....(From 3)

∆AOB=∆ BOC.....(Hypothenuse side test)

(2)To prove:

angle BCO= angle DCO

proof:

seg BO= seg DO....(4)

in ∆BOC,

seg BO = angle BCO....(5)

in ∆DOC,

seg DO=angle DCO.......(6)

from 4,5 and 6,

angle BCO= angle DCO

(3) diagonals bisect each other from 3

angle are congruent so diagonals bisect each other.

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