equilateral if median=altitude, <A=60^o (2012 From Tasks to Tasks, p9 Ukr)

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equilateral if median=altitude, <A=60^o (2012 From Tasks to Tasks, p9 Ukr)

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#1 h May 2, 2021, 6:59 PM

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In the triangle $ABC$ , the angle $A$ is equal to $60^o$ , and the median $BD$ is equal to the altitude $CH$ . Prove that this triangle is equilateral.

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#2 h May 2, 2021, 11:36 PM • 2 Y

Y by Jc426, mathleticguyyy

if CE is a median note CH<=CE. suppose angel B<angle C : then BD>CE>=CH, contradiction

so, angle B=C and equilateral

golry to god
tony BARKER

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#3 h May 3, 2021, 1:26 AM

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Connect $ED$ . Because $DA=DC$ and $\angle AED = 90^\circ$ , $\triangle ADE$ is equilateral. And $EC=DB$ , $\angle BED = 120^\circ = \angle EDC$ , so $\triangle EDB$ ≌ $\triangle DEC$ . Then $EB=DC$ , $AB=AC$ , and $\angle BAC = 60^\circ$ . So $\triangle ABC$ is equilateral.

This post has been edited 2 times. Last edited by Alan-Z, May 3, 2021, 1:28 AM
Reason: Have a mistake.

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