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.

erinna erinna · Answer 2 · 2019-12-02T08:05:22+01:00

Answer:

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.

[tex]Area(\triangle ACM)=Area(\triangle BCM)=\dfrac{Area(\triangle ABC)}{2}=\dfrac{x}{2}[/tex]

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

[tex]Area(\triangle ADC)=Area(\triangle ADM)=\dfrac{Area(\triangle ACM)}{2}=\dfrac{x}{4}[/tex]

[tex]Area(\triangle BCD)=Area(\triangle BMD)=\dfrac{Area(\triangle BCM)}{2}=\dfrac{x}{4}[/tex]

[tex]Area(\triangle ADC)=Area(\triangle BMD)=\dfrac{x}{4}[/tex]

Hence proved.