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

Open in App

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.

Suggest Corrections

221

Similar questions

Diagonal $A C$ of a parallelogram $A B C D$ bisects $∠ A$

∠ C

A B C D

Diagonal AC of a parallelogram ABCD bisects $∠ A$ (see Fig. below), Show that
(i) it bisects $∠ C$ also (ii) ABCD is a rhombus

Question 6 (i)
Diagonal AC of parallelogram ABCD bisects $∠$ A ( see the given figure). Show that
It bisects $∠$ C also,

Diagonal $A C$ of a parallelogram $A B C D$ bisects $∠ A$ (see the given figure). Show that

(i) It bisects $∠ C$ also.

(ii) $A B C D$ is a rhombus.

Question 6 (i)
Diagonal AC of parallelogram ABCD bisects $∠$ A ( see the given figure). Show that
It bisects $∠$ C also,

Join BYJU'S Learning Program