Question

m parallel to n, l is a transversal lines and AC and BD is angle bisector of angle 2 and angle 1 then prove to that AC parallel to BD​

Answer 1

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Step-by-step explanation:

∵l || m and AC is a transversal.

∴ ∠BAC = ∠DCA

[Alternate interior angles]

Also p || q and AC is a transversal,

∴ ∠BCA = ∠DAC

[Alternate interior angles]

Now, in ΔABC and ΔCDA,

∠BAC = ∠DCA

[Proved]

∠BCA = ∠DAC

[Proved]

CA = AC

[Common]

∴ Using ASA criteria, we have ΔABC ≌ ΔCDA

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