1 find $\triangle abe$ equal in area to quadrilateral $abcd$ in the figure below. to determine point $e$, draw in the following order and fill the appropriate words in the blanks. (1) draw diagonal $ac$. (2) draw a straight line passing through point $d$ and parallel to diagonal line $ac$. at this time, let $e$ be the intersection point with the extended line of $bc$, then connect points $a$ and $e$. (3) since straight line $de$ and diagonal $\square$ are parallel, $\triangle acd$ and $\triangle \square$ are equal. therefore, quadrilateral $abcd$ and $\triangle \square$ have the same area. 2 the following figure is trapezoid $abcd$ with $ad \parallel bc$. place point $m$ on side $bc$ and draw $am$ bisecting the area of trapezoid $abcd$. fill the appropriate words in the blanks. (1) place midpoint $e$ on side $dc$ and connect points $a$ and $e$. then let $f$ be the intersection point of the extended line of $ae$ and the extended line of $bc$. (2) a triangle with the same area as $\triangle ade$ is $\triangle \square$. (3) a triangle with the same area as trapezoid $abcd$ is $\triangle \square$. (4) let $m$ be the $\square$ of $bf$ and connect points $a$ and $m$. therefore, $am$ bisects the area of trapezoid $abcd$. 3 in the parallelogram in the following figures, points $p$, $q$, and $r$ are the midpoints of their respective sides. answer the ratio of the area of the shaded area to the area of parallelogram $abcd$. (1) (2) (ans.) : (ans.)

Question

Accepted Answer

### Problem 1
# Explanation:
## Step1: Analyze the parallel lines
From step (2), we know $ DE \parallel AC $. So the first blank should be $ AC $.
## Step2: Triangles with equal area
Triangles with the same base and between the same parallel lines have equal area. Since $ DE \parallel AC $, $ \triangle ACD $ and $ \triangle ACE $ (because they share the base $ AC $ and are between $ AC $ and $ DE $) have equal area. So the second blank is $ ACE $.
## Step3: Area of quadrilateral and triangle
Quadrilateral $ ABCD $ area is $ \triangle ABC + \triangle ACD $. Since $ \triangle ACD=\triangle ACE $, then $ ABCD $ area is $ \triangle ABC+\triangle ACE=\triangle ABE $. So the third blank is $ ABE $.

# Answer:
(3) $ AC $, $ ACE $, $ ABE $

### Problem 2
# Explanation:
## Step1: Triangle with same area as $ \triangle ADE $
Since $ AD \parallel BC $, $ \triangle ADE $ and $ \triangle CFE $ (vertical angles and parallel lines, so equal area). Wait, actually, $ E $ is the midpoint of $ DC $, $ AE $ extended meets $ BC $ extended at $ F $. $ \triangle ADE $ and $ \triangle FCE $ are congruent (ASA), so same area. So blank is $ FCE $.
## Step2: Triangle with area of trapezoid
Trapezoid $ ABCD $ area is $ \triangle ABC + \triangle ADC $. But with $ F $, trapezoid area is equal to $ \triangle ABF $ (since $ \triangle ADE=\triangle FCE $, so $ ABCD $ area = $ ABF $ area). So blank is $ ABF $.
## Step3: Midpoint of $ BF $
To bisect the area of $ \triangle ABF $, we take the midpoint of $ BF $, so $ M $ is the midpoint.

# Answer:
(2) $ FCE $
(3) $ ABF $
(4) midpoint

### Problem 3 (1)
# Explanation:
## Step1: Let area of parallelogram be $ S $
Let $ AB = a $, height $ h $, so $ S = ah $. $ P $ is midpoint of $ AB $, so $ BP=\frac{a}{2} $.
## Step2: Area of shaded triangle
Area of $ \triangle BPD $: base $ BP=\frac{a}{2} $, height $ h $ (same as parallelogram). Area $=\frac{1}{2}\times\frac{a}{2}\times h=\frac{1}{4}ah $? Wait, no, wait. Wait, in parallelogram $ ABCD $, $ P $ is midpoint of $ AB $, $ D $ is vertex. The area of $ \triangle BPD $: actually, $ \triangle ABD $ area is $ \frac{1}{2}S $. $ P $ is midpoint, so $ \triangle BPD $ area is $ \frac{1}{2}\times\frac{1}{2}S=\frac{1}{4}S $? Wait, no, maybe better: Let’s assume coordinates. Let $ A(0,0) $, $ B(b,0) $, $ D(0,h) $, $ C(b,h) $, $ P(\frac{b}{2},0) $. Area of $ \triangle BPD $: points $ B(b,0) $, $ P(\frac{b}{2},0) $, $ D(0,h) $. Area $=\frac{1}{2}\times base\times height=\frac{1}{2}\times\frac{b}{2}\times h=\frac{1}{4}bh $. Parallelogram area is $ bh $. Wait, no, that's wrong. Wait, $ \triangle BPD $: base $ BP = \frac{b}{2} $, height from $ D $ to $ AB $ is $ h $. Wait, no, $ D $ is at $ (0,h) $, $ AB $ is x - axis. So area of $ \triangle BPD $ is $ \frac{1}{2}\times BP\times h=\frac{1}{2}\times\frac{b}{2}\times h=\frac{bh}{4} $? But parallelogram area is $ bh $. Wait, no, maybe I messed up. Wait, $ P $ is midpoint of $ AB $, $ BD $ is diagonal? No, $ P $ is on $ AB $, $ D $ is opposite. Wait, the shaded triangle is $ \triangle BPD $? Wait, no, the figure: $ P $ on $ AB $, $ B $, $ P $, $ D $. Wait, actually, in parallelogram $ ABCD $, $ P $ is midpoint of $ AB $, so area of $ \triangle BPD $: $ AB $ is base, $ D $ is top. The area of $ \triangle ABD $ is $ \frac{1}{2} $ of parallelogram. $ P $ is midpoint, so $ \triangle BPD $ is $ \frac{1}{2} $ of $ \triangle ABD $, so $ \frac{1}{4} $ of parallelogram? Wait, no, $ \triangle ABD $ area is $ \frac{1}{2} \times AB \times height $. $ \triangle BPD $ has base $ BP = \frac{AB}{2} $, same height, so area $ \frac{1}{2} \times \frac{AB}{2} \times height=\frac{1}{4} \times AB \times height=\frac{1}{4} $ of parallelogram. Wait, but maybe another way: Let parallelogram area be $ 4 $ units. Then $ \triangle BPD $ area is $ 1 $? No, wait, maybe I made a mistake. Wait, the correct ratio: In parallelogram $ ABCD $, $ P $ is midpoint of $ AB $. The area of $ \triangle BPD $: $ AB $ length $ l $, height $ h $. Area of parallelogram $ lh $. Area of $ \triangle ABD $ is $ \frac{lh}{2} $. $ P $ is midpoint, so $ \triangle BPD $ area is $ \frac{1}{2} \times \frac{lh}{2}=\frac{lh}{4} $. So ratio is $ \frac{1}{4} $? Wait, no, wait the figure: $ P $ is on $ AB $, $ D $ is on $ AD $? No, $ ABCD $ is parallelogram, $ A - B - C - D - A $. $ P $ is midpoint of $ AB $, $ D $ is vertex. So $ \triangle BPD $: base $ BP $, height from $ D $ to $ AB $. So area $ \frac{1}{2} \times BP \times h $. $ BP=\frac{AB}{2} $, so $ \frac{1}{2} \times \frac{AB}{2} \times h=\frac{AB \times h}{4}=\frac{1}{4} $ of parallelogram area (since parallelogram area is $ AB \times h $).

## Step2: For the second figure (2)
Let $ AD = l $, height $ h $, parallelogram area $ lh $. $ Q $ is midpoint of $ AD $, $ R $ is on $ BC $. The shaded triangle: base $ QR $? Wait, $ Q $ is midpoint, $ R $ is... Wait, $ AQ=\frac{l}{2} $, the shaded triangle has base $ \frac{l}{2} $ and height $ h $? No, wait, the triangle is between $ A $, $ R $, and some point. Wait, actually, in parallelogram $ ABCD $, $ Q $ is midpoint of $ AD $, $ R $ is on $ BC $. The shaded triangle: area is $ \frac{1}{4} $ of parallelogram? Wait, no, let's take coordinates. $ A(0,0) $, $ B(b,0) $, $ C(b + l,h) $, $ D(l,h) $. $ Q $ is midpoint of $ AD $: $ (\frac{l}{2},h) $, $ R $ is midpoint of $ BC $: $ (b+\frac{l}{2},0) $. The shaded triangle: points $ A(0,0) $, $ R(b+\frac{l}{2},0) $, $ Q(\frac{l}{2},h) $. Area: $ \frac{1}{2} \times base \times height $. Base $ AR $? No, base is $ RQ $? Wait, no, the triangle is $ A $, $ R $, and the intersection? Wait, maybe easier: The area of the shaded triangle: since $ Q $ is midpoint of $ AD $, and $ R $ is such that... Wait, the parallelogram area is $ lh $ (length $ l $, height $ h $). The shaded triangle: base $ \frac{l}{2} $, height $ h $, area $ \frac{1}{2} \times \frac{l}{2} \times h=\frac{lh}{4} $? No, wait, maybe $ \frac{1}{4} $ for both? Wait, no, first figure: ratio $ \frac{1}{4} $, second figure: let's see, $ Q $ is midpoint of $ AD $, $ R $ is on $ BC $. The shaded triangle: area is $ \frac{1}{4} $ of parallelogram. Wait, maybe the first figure ratio is $ \frac{1}{4} $, second is $ \frac{1}{4} $? No, wait, first figure: $ P $ is midpoint of $ AB $, so $ \triangle BPD $ area is $ \frac{1}{4} $ of parallelogram. Second figure: $ Q $ is midpoint of $ AD $, the shaded triangle: area is $ \frac{1}{4} $ of parallelogram.

# Answer:
(1) $ \frac{1}{4} $
(2) $ \frac{1}{4} $

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QUESTION IMAGE

Question

Explanation:

Problem 1

Step1: Analyze the parallel lines

Step2: Triangles with equal area

Step3: Area of quadrilateral and triangle

Step1: Triangle with same area as \( \triangle ADE \)

Step2: Triangle with area of trapezoid

Step3: Midpoint of \( BF \)

Step1: Let area of parallelogram be \( S \)

Step2: Area of shaded triangle

Step2: For the second figure (2)

Answer:

Problem 2