Question

Can u pls solve this sum in a paper with clear explanation...Thank you..​

Answer 1

Answer:

Opposite angles of a parallelogram are equal

$\angle A = \angle C\\\\As \ AC \ bisects \ \angle A,\\\angle DAC = \angle BAC\\\\AB \parallel CD[opposte \ sides \ of \ a \ parallelogram \ are \,equal]\\\therefore \angle BAC = \angle ACD\\\\AD \parallel BC[opposte \ sides \ of \ a \ parallelogram \ are \,equal]\\\therefore \angle DAC = \angle ACB\\\\For \ \triangle ACB \ and \ \triangle ACD,\\\angle BAC = \angle ACD[proved]\\\angle DAC = \angle ACB[proved]\\AC =AC[common] \\\\\therefore \triangle ACB \cong \triangle ACD...(1)$$From \ this \ we \ can \ say \ that \ \\AB=AD[From \ (1)]\\BC=DC[From \ (1)]\\AB=CD[opposite \ sides \ of \ a \ parallelogram \ are \ equal]\\AD=BC[opposite \ sides \ of \ a \ parallelogram \ are \ equal]\\\therefore AB=BC=CD=AD\\\therefore ABCD \ is \ a \ rhombus \ whose \ sides \ are \ all \ equal \ but \ angles \ not$

Step-by-step explanation: