In trapezoid ABCD with bases BC and AD, diagonals AC and BD are drawn. Prove that S (ABD): S (BCD) = AD: BC ..

univerkov | April 19, 2021 | education

We will construct the height ВН to the base AD and the height DK for the continuation of the BC.

Since BC is parallel to AD, and BH and DK are perpendicular to them, then the quadrangle ВKDH is a rectangle, then BH = DK.

The area of the triangle ABD is equal to: Savd = AD * BН / 2 = AD * DK / 2.

The area of the triangle ВСD is equal to: Sвсд = ВС * DК / 2.

Then Savd / Svsd = ((AD * DK) / 2) / ((BC * DK) / 2) = AD / BC, as required.

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