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27 July, 17:07

The diagonal AC of a quadrilateral ABCD bisects angle BAD and angle BCD. Prove that BC=CD.

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  1. 27 July, 20:53
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    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
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