Equality of areas in a triangle

Show that the areas of the two triangles $ABM$ and $ACM$ , defined by a median of an arbitrary triangle $ABC$ as in the following figure, are equal.

I solved itMark it done Add to my listKeep it in your list

I like this problem4 likes

Liked by

Hints

Hint 1

Open this only if you want a small nudge before looking at the solutions.

Solutions

Reveal solutionsAre you sure? Give it a try first.

Solution by Ancient Tree

Discussions1 useful vote

By definition, the median of $ABC$ from vertex $A$ splits $BC$ into two segments of same length :
$BM=MC$ Furthermore, the two triangles $ABM$ and $AMC$ have the same height $AH$ .

The area of the two triangles are respectively $\frac{BM \times HA}{2}$ and $\frac{MC\times AH}{2}$ , which are therefore equal.

Solution by goldfinchFR

Discussions0 useful votes

Par définition, la médiane du triangle $ABC$ issue du sommet $A$ coupe $BC$ en deux segments de même longueur :
$BM=MC$ Par ailleurs, les deux triangles $ABM$ et $AMC$ ont la même hauteur $AH$ .

Les surfaces des deux triangles qui valent respectivement $\frac{BM \times HA}{2}$ et $\frac{MC\times AH}{2}$ sont donc identiques.

Report