In the quadrilateral ABCD, ∠ ADB = ∠ ABC = 105 ° and ∠ DAB = ∠ DCB = 45 ° prove: CD = ab

prove:
∵∠A+∠C+∠ABC+∠ADC=360º
That is 45 & # 186; + 45 & # 186; + 105 & # 186; + ADC = 360 & # 186;
∴∠ADC=∠ADB+∠BDC=165º
∠BDC=165º-∠ADB=60º
∵∠ABD=180º-∠A-∠ADB=180º-45º-105º=30º
∴∠DBC=105º-30º=75º
Make de bisection ∠ BDC, cross BC to E
Then ∠ BDE = ∠ CDE = 30 & # 186;
∵∠DEB=180º-75º-30º=75º
∴∠DEB=∠DBE=75º
∴BD=DE
And ∵ a = ∵ C = 45 & # 186;
∠ABD=∠CDE=30º
∴⊿ABD≌⊿CDE（AAS）
∴CD=AB

In the quadrilateral ABCD, ∠ ADB = ∠ ABC = 105 ° and ∠ DAB = ∠ DCB = 45 ° prove: CD = ab

In the quadrilateral ABCD, ∠ ADB = ∠ ABC = 105 ° and ∠ DAB = ∠ DCB = 45 ° prove: CD = ab

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