Diagonal AC of a parallelogram ABCD bisects angleA then show that it bisects angleC also

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Solution

In ΔADC and ΔCBA,

AD = CB
(Opposite sides of a ||gm)

DC = BA
(Opposite sides of a ||gm)

AC = CA (Common)

Therefore,
ΔADC ≅ ΔCBA
( by SSS congruence rule.)

∠ ACD = ∠ CAB ……(i)
(by CPCT)

∠ BCA = ∠ CAD…..(ii)
(By CPCT)

& ∠ CAB = ∠ CAD…..(iii)
(Given)

From eq i,ii,iii,
All 4 above angles are equal to each other..

Hence, ∠ ACD = ∠ BCA

Thus, AC bisects ∠C also.

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