In a triangle ABC, D is the mid point of BC. angADB=45 deg,angACB=30 deg, Find the value of angDAB.(without using Trigonometry)

Draw triangle ABC so that line segment BC is its base, bisected by point D (given). <ADC = 135 degrees, because it is supplemental to <ADB = 45 degrees. <DAC = 15 degrees so that triangle ADC = 180 degrees. Line segment DC across from 15-degree angle = 1/2 line segment AD across from 30-degree angle. So line segment BD (given = line segment DC) must be across from an angle in triangle ABD, which is 1/2 another angle in triangle ABD. <ABD = 90 degrees and <DAB = 45 degrees for triangle ABC to have <ACB = 30 degrees (given), <ABD = 90 degrees, and <BAC = 60 degrees (<DAB = 45 degrees + <DAC = 15 degrees) all add to 180 degrees. <DAB = 45 degrees. (Notice that line segment AB across from 30-degree angle in larger triangle = 1/2 line segment BC across from 60-degree angle in larger triangle.)

Mr taseb

I think, you are wrong. I don't understand your explanation. Acoording to my calculation
ang DAB = 30 deg.

dk_ch: I am Miss tsaeb. Anyway, I am rusty on the geometry and just posting my guess. How did you get to your conclusion?

sorry Miss taseb

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