Download Hints and Answers to HW 18

Homework 18 Hints and Answers
#2 Hints: Break the parallelogram up into triangles and use triangle congruence to get the desired
congruences.
#3 Hint: Use one of the diagonals to break the quadrilateral into congruent triangles.
#4 Hint: Create an isosceles triangle and use that to help you get the angle congruence you need.
#7 Hint: Pick two points on m, drop perpendiculars to l, and use the resultant parallelogram to
prove what you want.
#9 Hint: Pick a point on m and drop a perpendicular to l. Show the perpendicular is perpendicular
to both l and m.
2. Part 1: Because 𝐴𝐷 ∥ 𝐵𝐶, then the alternate interior angles
created by the transversal 𝐴𝐶 must be congruent (Converse to
the Alt Int Ang Thm). So ∠1 ≅ ∠3. Likewise, 𝐴𝐵 ∥ 𝐶𝐷
A
B
∠1 ∠4
implies that ∠2 ≅ ∠4. So by ASA, ∆𝐴𝐵𝐶 ≅ ∆𝐶𝐷𝐴. We can
show that ∆𝐴𝐵𝐷 ≅ ∆𝐶𝐷𝐵 using a similar proof.
∠3
Part 2: Because ∆𝐴𝐵𝐶 ≅ ∆𝐶𝐷𝐴, we get 𝐴𝐵 ≅ 𝐶𝐷 and
∠2
D
𝐴𝐷 ≅ 𝐵𝐶.
C
Part 3: Because ∆𝐴𝐵𝐶 ≅ ∆𝐶𝐷𝐴, the opposite angles
∠ABC and ∠CDA are congruent. Because ∆𝐴𝐵𝐷 ≅ ∆𝐶𝐷𝐵, the opposite angles ∠BAD and
∠DCB are congruent.
Part 4: Since ABCD is a parallelogram, it is convex.
Therefore, the diagonals intersect in their interiors. Let E be
the point of intersection. Note that because 𝐴𝐵 ∥ 𝐶𝐷, we
A
B
∠8 ∠7
know that ∠5 ≅ ∠7 and ∠6 ≅ ∠8. We also know that
𝐴𝐵 ≅ 𝐶𝐷, so ∆𝐴𝐵𝐸 ≅ ∆𝐶𝐷𝐸 (ASA). This implies that AE
E
= EC and DE = EB. Then E is the midpoint of both 𝐴𝐶 and
∠5
∠6
D
𝐷𝐵.
C
3. In order to show that ABCD is a parallelogram, we need
to show that the pairs of sides that are across from each other
are parallel. We already know that 𝐴𝐵 ∥ 𝐶𝐷, so we just need
to show that 𝐴𝐷 ∥ 𝐵𝐶. Consider the triangles ABC and
CDA. We know that ∠1 ≅ ∠2 (Converse of Alt Int Ang
Thm) and 𝐴𝐵 ≅ 𝐶𝐷 (given), so these two triangles are
congruent (SAS). Consequently, ∠3 ≅ ∠4. But ∠4 ≅ ∠5
(Converse of Alt Int Ang Thm). It follows that ∠3 ≅ ∠5,
which implies that 𝐴𝐷 ∥ 𝐵𝐶 (Corresponding Angles Thm).
A
∠5
∠4
D
∠1
∠3
∠2
C
B
4. Because 𝐴𝐵 ≅ 𝐴𝐷, ABD is isosceles. So ∠𝐴𝐵𝐷 ≅ ∠𝐴𝐷𝐵. Also,
because 𝐴𝐵 ∥ 𝐶𝐷, we know that ∠𝐴𝐵𝐷 ≅ ∠𝐵𝐷𝐶 (Conv Alt Int Ang
Thm). This implies that ∠𝐴𝐷𝐵 ≅ ∠𝐵𝐷𝐶. Furthermore, because
trapezoids are convex (result proven in class), B is in the interior of
∠ADC. So 𝐷𝐵 is the angle bisector for ∠ADC.
B
A
D
m
l
P
P'
Q
Q'
C
7.
Let l and m be two parallel lines, and P,
Q be distinct points on m. Drop
perpendiculars from P and Q to l, and let the feet be P' and Q',
respectively. The quadrilateral PQQ'P' has two right angles and two
parallel sides. Furthermore, because all four angles at P' and Q' are
right angles. Then 𝑃𝑃′ and 𝑄𝑄′ are parallel (Alt Int Ang Thm). So
PQQ'P' is a parallelogram, which implies that 𝑃𝑃′ ≅ 𝑄𝑄′. This in
turn implies that d(P, l) = d(Q, l).
9. Let l and m be two parallel lines. Choose P on m and drop a perpendicular to l. Call this
perpendicular t. Note that at the intersection of t and l, all four angles are right angles. Since l
and m are parallel, then the interior angles at the intersection of t and m must also be right
angles (Conv Alt Int Ang Thm). So t is also perpendicular to m.