Prove Triangle ABC is an isosceles
Well first you need to know what makes a triangle isosceles in terms of angles. Then you'd wanna look at the angles using rules for the angles inside triangles.
but how will I make use of the special segments to prove abc is an isosceles? :(
Segments inside the triangle? Well, you know the ones with those small boxes are 90 degrees. And the total interior angle of a triangle is....? And I see you have those equal signs on one of the lines inside the triangle. You should have another one on another line segment, signifying that they are of equal lengths. Triangles with same sides have the same angles.
i need to write a two-column proof for this and I still dont know how to get started :(
start wid whats given,
and use @wolfe8 steps, prove the base angles are congruent first, then conclude sides opposite to them are congruent
give it a try...
Just always remember, the prrof on isoceles triangles is that it has 2 or more sides congruent and therefore, its base angles must be equal as well :)
But I cant use ITT as a reason. Probably I should use CPCTC for this :(
yes ! u must use CPCTC to prove base angles are congreuent, put down statements first... then u can tweak them :)
hey @kelceyroche , maybe you can isolate the two right triangles there and you can prove that they are congruent, with than you can prove that the triangle is isosceles
oh yes got it, so triangle ACD is congruent to triangle ABE by AAS theorem then Angle B is Congruent to angle C by CPCTC? :D
Although I do think your drawing is missing something. I can't tell (although from experience I can guess) which lines have the same length.
there are Given statements, below the drawing, which say altitudes are congruent @wolfe8
and then Angle D is congruent to Angle E by perpendicularity?
yes both are right angles, so they're equal
can I use perpendicularity as a reason? :)
I am pretty sure the proof I outlined in the previous post is what they want you to do.
you familiar wid Leg Hypotenuse theorem ?
have a look at phi's pervious post :)
Do i still lack anything in my two-column proof?
@ganeshie8
yes, its bit off on 5th row,
what else do I lacK? :(
1. \(\large CD \perp AB \) 1. Given 2. \(\large BE \perp AC \) 2. Given 3. \(\large BE \cong CD \) 3. Given 4. \(\large \triangle DBC \cong \triangle ECB \) 3. By Leg Hypotenuse congruence
after that, you use CPCTC and to justify base angles are congruent
If you use the hypotenuse-leg theorem, then that shows the two right triangles are congruent. You should put in a line saying you have a right triangle
1. \(\large CD \perp AB \) 1. Given 2. \(\large BE \perp AC \) 2. Given 3. \(\large BE \cong CD \) 3. Given 4. \(\large BC \cong CB \) 3. By reflexive property 5. \(\large \triangle DBC \cong \triangle ECB \) 3. By Leg Hypotenuse congruence
yes, add that statement also which explicitly says D and E are right angles
how am i suppose to reason that they are right angles? @phi @ganeshie8
perpendicular BE forms a right angle
Thank you so much!!! and then I end with the SSS postulate?
1. \(\large CD \perp AB \) 1. Given 2. \(\large \angle BDC \cong 90 \) 2. definition of perpendicular 3. \(\large BE \perp AC \) 3. Given 4. \(\large \angle BEC \cong 90 \) 4. definition of perpendicular 5. \(\large BE \cong CD \) 5. Given 6. \(\large BC \cong CB \) 6. By reflexive property 7. \(\large \triangle DBC \cong \triangle ECB \) 7. By Leg Hypotenuse congruence
SSS must never enter this proof, your next step should be CPCTC
sss is the "long way" to show 2 right triangles are congruent. the hypotenuse-leg theorem is the short way: if you have a right triangle, and the hypotenuse and leg are congruent to another right triangle, the triangles are congruent. now you can do CPCT to show the base angles ABC and ACB are congruent. and then two congruent base angles mean you have an isosceles triangle
you would use SSS if you did not know the hypotenuse - leg theorem Given: hypotenuse and one leg congruent with those of another right triangle, you can say by pythagoras that the 3rd side is congruent, and so have 3 congruent sides., so by sss you have congruent triangles.
here is the complete proof, see if it makes sense 1. \(\large CD \perp AB \) 1. Given 2. \(\large \angle BDC \cong 90 \) 2. definition of perpendicular 3. \(\large BE \perp AC \) 3. Given 4. \(\large \angle BEC \cong 90 \) 4. definition of perpendicular 5. \(\large BE \cong CD \) 5. Given 6. \(\large BC \cong CB \) 6. By reflexive property 7. \(\large \triangle DBC \cong \triangle ECB \) 7. By Leg Hypotenuse congruence 8. \(\large \angle DBC \cong \angle ECB \) 8. By CPCTC 9. \(\large AB \cong AC \) 9. Converse of isosceles triangle theorem 10. \(\large \triangle ABC \) is isosceles 10. By definition of isosceles triangle (two sides congruent)
hank you so much!! we didn't study HL yet but now I can explain it, haha thank you so so so so much!!
np :) if you didnt study HL, better take it all the way to SSS or SAS... or watever u familiae wid :)
But I think our teacher gives these homeworks bc she wants us to read about all these in advance :D so atleast I can be active in class tomw :D
that sound good :) but again, if u want to go SSS way, u just need to include two more statements between 6th and 7th statements above
so how Do i do it SSS?
1. \(\large CD \perp AB \) 1. Given 2. \(\large \angle BDC \cong 90 \) 2. definition of perpendicular 3. \(\large BE \perp AC \) 3. Given 4. \(\large \angle BEC \cong 90 \) 4. definition of perpendicular 5. \(\large BE \cong CD \) 5. Given 6. \(\large BC \cong CB \) 6. By reflexive property 6.i. \(\large CD \cong BE \) 6.i. By pythagorean theorem, if two leg pairs are same, third leg pair wud be same as well 7. \(\large \triangle DBC \cong \triangle ECB \) 7. By SSS congruence 8. \(\large \angle DBC \cong \angle ECB \) 8. By CPCTC 9. \(\large AB \cong AC \) 9. Converse of isosceles triangle theorem 10. \(\large \triangle ABC \) is isosceles 10. By definition of isosceles triangle (two sides congruent)
thats SSS. Enjoy.. :)
thank you so much!!! :D
yw :)
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