In rectangle ABCD, O is the intersection point of the diagonals AD and BC, BH and DE are the heights of triangles

| June 11, 2021 | education

In rectangle ABCD, O is the intersection point of the diagonals AD and BC, BH and DE are the heights of triangles ABO and COD, respectively. BOH is 60 degrees, AH = 5 cm. Find OE.

In order to solve this problem, you need to carefully study the condition and draw a picture to facilitate understanding of the situation.

<HBO = 30 ° (180 is BHO angle).
BO = AO (by condition), therefore, <ОВА and <ОАВ are equal.
<BOA = 60 °. Since <ОВА = <ОАВ, <ОВА = (180 – 60) / 2 = 60 °.
<BOA = <OBA = <OAB = 60 °.
This means that triangle AOB is equilateral.
VN – height, bisector, median.
Since AH = 5, BH is the median, then AH = HO = 5cm.
AO = CO = 5 + 5 = 10 cm.
Likewise, the triangle is CDO-equilateral.
DE – median, OE = 10/2 = 5 cm.

Therefore, our answer is 5 cm.



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