The diagonal ac of a quadrilateral abcd bisects angle bad and angle bcd. prove that bc=cd.

BC = DC ⇒ proved

Step-by-step explanation:

* Lets explain how to solve the problem

- ABCD is a quadrilateral

- Its diagonal AC bisects angles BAD and BCD

- That means it divides the angle into two equal parts

∵ AC bisects angle BAD

∴ m∠BAC = m∠DAC

∵ AC bisects angle BCD

∴ m∠BCA = m∠DCA

* Lets revise a case of congruent we will use it to solve the problem

- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ  

 ≅ 2 angles and the side whose joining them in the 2nd Δ

- In ΔABC and ΔADC

∵ m∠BAC = m∠DAC ⇒ proved

∵ m∠BCA = m∠DCA ⇒ proved

∵ AC is a common side in both triangles

∴ ΔABC ≅ ΔADC ⇒ by ASA

- From congruent

∴ AB = AD

∴ BC = DC ⇒ proved

the answer is somewhere there guys it really . there im just not gonna answer

step-by-step explanation:

but i'm guessing ur first option tho

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step-by-step explanation: