In △abc, point m is the midpoint of side ab and point d is the midpoint of segment mc . prove that aadc=abmd.

A line segment from a vertex to the midpoint of the opposite side is a "median". A median divides the area of the triangle in half, as it divides the base in half without changing the altitude.

AAMC is half AABC. AADC is half AAMC, so is 1/4 of AABC. (By the formula for area of a triangle.)

ABMC is half AABC. ABMD is half ABMC, so is 1/4 of AABC. (By the formula for area of a triangle.)

Then, AADC = 1/4 AABC = ABMC, so AADC = ABMC by the transitive property of equality.

Given: In △ABC, point M is the midpoint of side AB and point D is the midpoint of segment MC .

Prove: Area(ADC)=Area(BMD).

Each median of a triangle divides the area of the triangle in two equal parts.

Let the area of triangle ABC is x square units.

Point M is the midpoint of side AB. It means MC is the median.

Point D is the midpoint of segment MC. So, AD and BD are medians of △ACM and △BCM respectively.

Hence proved.

answer: y=4000x+14000

step-by-step explanation:

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)