Question

Two line segments AB and CD bisect each other at O.
Prove that
(i) AC = BD (ii) angleCAB = angleABD
(iii) AD || CB (iv) AD=CB.

pls help!
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Answer 1

(i) In ∆ AOC and ∆BOD we have,

OC=OD (GIVEN)

OA=OB (GIVEN)

ANGLE COA =ANGLE BOD

(VERTICALLY OPPOSITE ANGLES)

SO ∆AOC==~(CONGRUENT)∆BOD

AC=BD by (C.P.C.T.)

(ii)ANGLE CAB=ANGLE ABD by (C.P.C.T.)....

(iii) AS WE KNOW THE QUADRILATERAL WHOSE DIAGONALS BISECT EACH OTHER IS A PARALLELOGRAM.....

SO ABCD IS A PARALLELOGRAM

THEN, AB ll BC (PROVED ABOVE)

(iv)AD=CB (OPPOSITE SIDES OF A PARALLELOGRAM ARE EQUAL)

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pLeAse

Answer 2

Answer:

AB and CD bisect each other at O i.e, AO=BO and CO=DO

in ΔCOA and ΔDOB

Given CO=OD,∠COA=∠BOD [ vertically opp angles]

AD=BD

∴ΔCOA≅ΔBOD

(i) ∴AC=BD[C.P.CT]

(ii) ∠CAB=∠ABD[C.P.CT]

again

in ΔCOB and ΔAOD

CO=OD [given]

BO=AO [given]

∠COB=∠AOD [vertically opp angles]

∴ΔCOB≅ΔAOD

∴∠CBA=∠BAD [ C.P. C.T]

(iii) and so AD∣∣CD [ ∵∠CBA=∠BAD which are altanate angles]

and AD=CB [C.P.C.T]

Step-by-step explanation:

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