Geometry Proofs (2 Viewers)

[Smile12345]

of course!

Very good!!! Ok... You sure??

 

[HeroicPandas]

Very good!!! Ok... You sure??

yes 100% correct! perfect working out, nothing wrong! u may consider my hint of extending line CD, it will save u time and will give u some experience on constructing lines

 

[Smile12345]

How about this one....



Given info: Triangle ABC is isosceles with AB = AC. D and E are points on BC such that Ang DAB = Ang CAE.

Prove that Triangle ADE is isosceles...

My answer;
In Triangles ABD, AEC,
AB = AC (Given) - Corresponding sides are equal)
Ang BAD = Ang CAE (Given)
Ang ABD = Ang ACE ( Triangle ABC is isosceles)
Therefore Triangle ABC is congruent to Triangle AEC (AAS)

AD = AE (Corresponding sides in congruent triangles)
Therefore Ang ADE = Ang AED

Therefore Triangle ADE is isosceles...

Am I on the right track anyone? Thanks in advance.

 

[Smile12345]

yes 100% correct! perfect working out, nothing wrong! u may consider my hint of extending line CD, it will save u time and will give u some experience on constructing lines

Very good... Thanks heaps... Yes, I should have considered your hint, it's definitely a lot quicker!! How about my next question?

 

[bedpotato]

How about this one....

View attachment 28356

Given info: Triangle ABC is isosceles with AB = AC. D and E are points on BC such that Ang DAB = Ang CAE.

Prove that Triangle ADE is isosceles...

My answer;
In Triangles ABD, AEC,
AB = AC (Given) - Corresponding sides are equal)
Ang BAD = Ang CAE (Given)
Ang ABD = Ang ACE ( Triangle ABC is isosceles)
Therefore Triangle ABC is congruent to Triangle AEC (AAS)

AD = AE (Corresponding sides in congruent triangles)
Therefore Ang ADE = Ang AED

Therefore Triangle ADE is isosceles...

Am I on the right track anyone? Thanks in advance.

Yep, you're right. But after the bold part, you can write something like " therefore triangle ADE is isosceles because it has two equal sides, and the angles opposite these sides (the base angles) are equal."

 

[Smile12345]

Yep, you're right. But after the bold part, you can write something like " therefore triangle ADE is isosceles because it has two equal sides, and the angles opposite these sides (the base angles) are equal."

Thanks heaps... Sounds good... I'll add that in.

 

[Smile12345]

Is the following correct...?

Diagram:

Question: Triangle ACD is isosceles, AC = DC. Triangle ABC is equilateral. Prove the AB is perpendicular to AD.

AB = BC = AC = DC

Ang BAC = Ang ABC = Ang ACB = 60 degrees (Angles in equilateral triangle)
Ang DAC + Ang BCA = 180 degrees (Straight line BCD)
Therefore Ang DCA = 120 degrees.
Ang CAD + Ang CDA = 180-120/2 = 30 degrees (Triangle ACD is isosceles)

Ang ABC + Ang DAC = 90 degrees
60 deg + 30deg = 90 deg
Therefore AB is perpendicular to AD

 

[bedpotato]

Yep, that's right.

 

[Smile12345]

Yep, that's right.

Very good! Is there anywhere I could improve it?

The next one I'm not sure about.... Could you help me if I put it up?

 

[Smile12345]

Could someone please check what I have is right?? And then where to go next...

Given: ABCD is a parallelogram, BX=DY

Diagram:

Question a) Prove Triangle ABX is congruent to CDY
b) Hence show AX = CY
c) Prove that AXCY is a parallelogram.

a) In triangles ABX, CDY
BX = DY (Given)
AB = DC (Opp. sides are = in parallelogram)
Angle ABX = Angle CDY (Opposite angles are = in parallelogram...
Therefore Triangle ABX is congruent to Triangle CDY

b) AX= CY (Corresponding sides on congruent triangles)... Do I need more here?

c) Not sure...

By the way - Does anyone use the abbreviation Parm. for Parallelogram?

Thanks for your help in advance.

 

[bedpotato]

Could someone please check what I have is right?? And then where to go next...

Given: ABCD is a parallelogram, BX=DY

Diagram: View attachment 28358

Question a) Prove Triangle ABX is congruent to CDY
b) Hence show AX = CY
c) Prove that AXCY is a parallelogram.

a) In triangles ABX, CDY
BX = DY (Given)
AB = DC (Opp. sides are = in parallelogram)
Angle ABX = Angle CDY (Opposite angles are = in parallelogram...
Therefore Triangle ABX is congruent to Triangle CDY (SAS)

b) AX= CY (Corresponding sides on congruent triangles)... Do I need more here?

c) Not sure...

By the way - Does anyone use the abbreviation Parm. for Parallelogram?

Thanks for your help in advance.

a) That's right, but you have to write the reason at the end. Why is it congruent? Is AAA, AAS, SAS? Well, in this case it's SAS. Other than that, a is fine.
b) That's right. You don't ave to add anything else.
c) Okay, here you have to prove that angle YAX = angle YCX, and angle AXC = angle AYC. You've already proved that AX = CY, so all you have to prove now is AY = XC. Try it again and see if you can get the answer.

I don't use parm., I just write parallelogram.

 

[Smile12345]

a) That's right, but you have to write the reason at the end. Why is it congruent? Is AAA, AAS, SAS? Well, in this case it's SAS. Other than that, a is fine.
b) That's right. You don't ave to add anything else.
c) Okay, here you have to prove that angle YAX = angle YCX, and angle AXC = angle AYC. You've already proved that AX = CY, so all you have to prove now is AY = XC. Try it again and see if you can get the answer.

I don't use parm., I just write parallelogram.

Good, I got a and b right!! Ok I'll try c again....

I wonder ... Does anyone else use parm.??? Do you reckon it would be ok??

Thanks for your help.

 

[bedpotato]

Good, I got a and b right!! Ok I'll try c again....

I wonder ... Does anyone else use parm.??? Do you reckon it would be ok??

Thanks for your help.

And you're welcome.

 

[Smile12345]

Well, it's an abbreviation for parallelogram, so I think so, but I'm not fully sure.
And you're welcome.

Yes, it's definitely an abbreviation for parallelogram, I was just wondering if anyone else used it?!!

I do appreciate it... I've thought about part c, but I'm not exactly sure how to set this out, could someone please help me out?...

 

[bedpotato]

c) Okay, here you have to prove that angle YAX = angle YCX, and angle AXC = angle AYC. You've already proved that AX = CY, so all you have to prove now is AY = XC. Try it again and see if you can get the answer.

I've thought about it but I'm not exactly sure how to set this out, could someone please help me out...

Okay, I'm going to write ^ instead of angle.

Firstly:

^AXB = ^DYC (corresponding <'s of congruent triangles)
Let <AXB = x and ^DYC = x
^AXC = 180 - ^AXB (supplementary ^'s)
= 180 - x
^AYC = 180 - ^DYC (supplementary ^'s)
= 180 - x
Therefore:
^AYC = ^AXC = 180 -x

Secondly:
^BAD = ^DCB (opp. ^'s of parallelogram)
Let ^BAD =k and ^DCB = k

and ^BAX = ^DCY
Let ^BAX = h and ^DCY = h

^YAX = ^BAD - ^BAX
= k - h
^YCX = ^DCB - ^DCY
= k - h
Therefore:
^YAX = ^YCX = k - h

^AYC = ^AXC and ^YAX = ^YCX
Therfore, AYXC is a paralellogram (2 pairs of opposite angles are equal)

You can use the sides instead of the angles though.

 

[Smile12345]

Okay, I'm going to write ^ instead of angle.

No worries... Much better idea... Thanks heaps...

Yeah ok then... So if I did the sides, I could say AX = CY (Proven in b) and then AY=XC (Opposite sides are equal - parallelogram property)... Or isn't this enough detail?

 

[bedpotato]

Okay, I'm going to write ^ instead of angle.

No worries... Much better idea... Thanks heaps...

Yeah ok then... So if I did the sides, I could say AX = CY (Proven in b) and then AY=XC (Opposite sides are equal - parallelogram property)... Or isn't this enough detail?

Not quite...

The first part is right, AX = CY
But for the second part, you'll have to do something like this:

AD = BC (opp. sides of parallelogram are equal)
So, let AD and BC = a
BX = YD (corresponding sides of congruent triangles)
So, let BX and YD = b

AY = AD - DY
= a - b
XC = BC - BX
= a - b
Therefore, AY = XC = a - b

AY = XC and AX = YC
Therefore, AYCX is a parallelogram (2 pairs of opp. sides are equal)

 

[Smile12345]

Not quite...

The first part is right, AX = CY
But for the second part, you'll have to do something like this:

AD = BC (opp. sides of parallelogram are equal)
So, let AD and BC = a
BX = YD (corresponding sides of congruent triangles)
So, let BX and YD = b

AY = AD - DY
= a - b
XC = BC - BX
= a - b
Therefore, AY = XC = a - b

AY = XC and AX = YC
Therefore, AYCX is a parallelogram (2 pairs of opp. sides are equal)

Right ok... Thanks very much...

 

[Smile12345]

Thanks all for your help, it is greatly appreciated...

 

[bedpotato]

You're welcome