Prove that the parallelogram $ABCD$ with relation $\angle ABD + \angle DAC = 90^o$, is either a rectangle or a rhombus.
Problem
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Tags: geometry, parallelogram, rectangle, rhombus
ss28112
25.03.2020 20:19
Angle chasing yields us with the fact that the diagonals must be perpendicular, leaving with a rhombus. I don't think a rectangle works unless it is a square.
brainiacmaniac31
25.03.2020 21:43
ss28112 wrote:
Angle chasing yields us with the fact that the diagonals must be perpendicular, leaving with a rhombus. I don't think a rectangle works unless it is a square.
I don't think your solution is correct, because a rectangle works but its diagonals aren't perpendicular.
ss28112
25.03.2020 22:32
Oops I labeled an angle wrong.
sunken rock
25.03.2020 22:53
Just apply symmedian property! Best regards, sunken rock
AdventuringHobbit
26.03.2020 01:22
Assume ABCD is not a rectangle. We find from the condition that DAC = 90 - ABD, hence AC passes through the circumcenter of ABD. Since it also passes through the midpoint of BD, ABD is isosceles with AB=AD as desired.
Kennigsta
27.03.2020 22:28
The question is not correct well. $Proof:$ Suppose, $\angle ABD = 70^o$, $\angle DAC = 20^o$. It can be proven that $\angle BAC+ \angle DBC=90^o$ too. Now suppose that $ \angle BAC = 30^o$ and $\angle DBC = 60^o$. Then it's still parallelogram but the relation holds. $\Box$