In the diagram below triangle ABE is an isosceles right angle triangle and triangle ADC is an isosceles right angel triangle and M is in the middle of BC
prove that triangle EMD is an isosceles right angel triangle
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Batominovski
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1Do you mean that $\triangle A\color{red}BE$ is an isosceles right triangles? – Dr. Mathva May 03 '19 at 13:34
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1Median is for a triangle, not for a line segment – Tojra May 03 '19 at 13:55
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Let $L$ be a midpoint of $AB$, and $K$ be a midpoint of $AC$.
$EL=\frac{1}{2}AB=MK$
$ML=\frac{1}{2}AC=DK$
$\angle ELM=\angle BLM + 90^{\circ}=\angle MKC+90^{\circ}=\angle MKD$
By SAS congruence criterion $\triangle ELM\equiv \triangle MKD$
From congruence we get $EM=DM$ and $\angle EML=\angle MDK$.
$\angle EMD=\angle EML+\angle LMK+\angle KMD=\angle MDK+ \angle MKC+\angle KMD=180^{\circ}-\angle CKD=90^{\circ}$
Kulisty
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