Given that ABCD is a parallelogram with diagonals intersecting at M. Segment PQ is drawn such that P lies on AB and Q lies on DC and it passes through M. Prove that PM is congruent to QM.
Answers
Answer:
we thus have enough informnation to prove that the two triangles are congruent. Because of thism, one side of the triangles, Pm which is the congruent side to that of PQ must be equal as well and
PM = QM
Step-by-step explanation:
because it's a parrallelogram, Ad = BC and AB = DC.
P passes through both AB and DC on a diagonal, angles created by P have oppoiste exterior angles of eachother.
the diagonals of AC create congruent angles at DMC and AMB and because P is cutting through them, the angles P cuts at DMQ and QMC are equal to that of AMP and PMB
that being said, we now see that the angles of APM triangle and AMC triangol are equal and because the diagonal of BD is is being cut by AC, which AD is parralel to BC and AB to Dc, we know now that lines AM and MC are congruent
we thus have enough informnation to prove that the two triangles are congruent. Because of thism, one side of the triangles, Pm which is the congruent side to that of PQ must be equal as well and
thus PM = QM
The alternate interior angles formed by the parallel sides and a common
transversal are equal, such that ΔPBM and ΔQDM are congruent.
Correct response;
PM is congruent to QM by CPCTCHow is the congruency of the segments determined?The two column proof is presented as follows;
Statement [tex]{}[/tex] Reason
ABCD is a parallelogram [tex]{}[/tex] Given
AC and DB are diagonals of ABCD [tex]{}[/tex] Given[tex]{}[/tex]
M is the point of intersection of the diagonals [tex]{}[/tex] Given
MB ≅ DM, and AM ≅ CM;[tex]{}[/tex] Properties of the diagonals of a parallelogram
∠PMB ≅ ∠DMQ [tex]{}[/tex] Vertical angles theorem
∠PBM ≅ ∠QDM [tex]{}[/tex] Alternate interior angles theorem
ΔPBM ≅ ΔQDM [tex]{}[/tex] Angle-Side-Angle Congruency post.
PM ≅ QM [tex]{}[/tex] CPCTC
ΔPBM is congruent to ΔQDM by Angle-Side-Angle congruency
postulate, given that they have two angles and an included side that are
congruent.
Therefore;
PM is congruent to QM by Corresponding Parts of Congruent Triangles are Congruent, CPCTC.Learn more about congruency postulates here:
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I hope this helped! :)
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The first step that we must take before attempting to solve almost any problem is to determine what the problem statement is asking us to do and what they are giving to us to help finish the problem. Looking at this question we can see that they have requested for us to solve for x and we were provided with a figure.
After understanding what the problem is asking for us to do we can attempt to solve the problem. We can see that there are two angles on a straight line and this is known as supplementary angles. Let us define what supplementary angles are.
Supplementary Angles ⇒ Supplementary Angles are two angles that are formed on a straight line causing them to add up to 180 degrees. Therefore, an expression can be made where the two angles add up to 180 and it can be solved for the unknown.In this case we don't have a missing angle but rather a variable that makes up each of the angles. Using a similar concept we can create an expression with 4x - 6 and 2x and have them combined to equal to 180. After that we can solve for the unknown which in turn would give us x.
Create an expression
[tex]Angle_1 + Angle_2 = 180[/tex][tex](2x) + (4x - 6) = 180[/tex]Combine like terms
[tex]2x + 4x - 6 = 180[/tex][tex](2x + 4x) - 6 = 180[/tex][tex](6x) - 6 = 180[/tex]Add 6 to both sides
[tex](6x) - 6 + 6 = 180 + 6[/tex][tex](6x) = 180 + 6[/tex][tex]6x = 186[/tex]Divide both sides by 6
[tex]\frac{6x}{6} = \frac{186}{6}[/tex][tex]x = \frac{186}{6}[/tex][tex]x = 31[/tex]Therefore, after creating an expression and then simplifying it down all the way to just have the value of x equal to something we were able to determine that the unknown variable of x is equal to 31.
[tex]\huge\text{Hey there!}[/tex]
[tex]\text{Your equation: }\rm{(2x) + (4x - 6) = 180^\circ}\\\large\text{SOLVING for the x-value}\\\rm{2x + 4x - 6 = 180^\circ}\\\large\text{COMBINE the LIKE TERMS}\\\rm{(2x + 4x) + (-6) = 180^\circ}\\\rm{2x + 4x - 6 = 180^\circ}\\\rm{6x - 6 = 180^\circ}\\\large\text{ADD 6 to BOTH SIDES}\\\rm{6x - 6 + 6 = 180 + 6}\\\large\text{SIMPLIFY IT!}\\\rm{6x = 180 + 6}\\\rm{6x = 186}\\\large\text{DIVIDE 6 to BOTH SIDES}\\\rm{\dfrac{6x}{6} = \dfrac{186}{6}}\\\large\text{SIMPLIFY IT!}\\\rm{x = \dfrac{186}{6}}\\\rm{x= 31}[/tex]
[tex]\huge\text{Therefore, your answer should be: \boxed{\mathsf{x = 31}}}\huge\checkmark\\\\\\\\\huge\text{Good luck on your assignment \& enjoy your day!}\\\\\\\\\frak{Amphitrite1040:)}[/tex]
can someone answer this asap :))
^^ all i need is the second one shown in the picture, you don’t have to do #1 :)
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HAVE A GOOD DAY!
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