Mixing
Proving that a triangle is isosceles.
In a triangle $ABC$, let $D$ be a point on the segment $BC$ such that $AB + BD = AC + CD$. Suppose that the points $B,C$ and the centroids of triangles $ABD$ and $ACD$ lie on a circle. Prove that $AB = AC$.
I began this way: Denote the centroid of triangle $ABD$ by $G_1$ and the centroid of triangle $ACD$ by $G_2$. Then by using $Basic$ $Proportionality$ $Theorem$, I got that $BG_1G_2C$ is an isosceles trapezium. After that I am stuck. Please help.
geometry triangle
edited Jan 1 at 8:07
Michael Rozenberg
97.5k1589188
asked Jan 1 at 6:11
Anu RadhaAnu Radha
1359
add a comment |
In a triangle $ABC$, let $D$ be a point on the segment $BC$ such that $AB + BD = AC + CD$. Suppose that the points $B,C$ and the centroids of triangles $ABD$ and $ACD$ lie on a circle. Prove that $AB = AC$.
I began this way: Denote the centroid of triangle $ABD$ by $G_1$ and the centroid of triangle $ACD$ by $G_2$. Then by using $Basic$ $Proportionality$ $Theorem$, I got that $BG_1G_2C$ is an isosceles trapezium. After that I am stuck. Please help.
geometry triangle
edited Jan 1 at 8:07
Michael Rozenberg
97.5k1589188
asked Jan 1 at 6:11
Anu RadhaAnu Radha
1359
Try to prove that BD=CD
– Moti
Jan 1 at 7:08
Yup, I tried that, too, @Moti. But I feel I am missing something which I could gather from the given information (or by any other manipulation)
– Anu Radha
Jan 1 at 7:39
add a comment |
In a triangle $ABC$, let $D$ be a point on the segment $BC$ such that $AB + BD = AC + CD$. Suppose that the points $B,C$ and the centroids of triangles $ABD$ and $ACD$ lie on a circle. Prove that $AB = AC$.
I began this way: Denote the centroid of triangle $ABD$ by $G_1$ and the centroid of triangle $ACD$ by $G_2$. Then by using $Basic$ $Proportionality$ $Theorem$, I got that $BG_1G_2C$ is an isosceles trapezium. After that I am stuck. Please help.
geometry triangle
edited Jan 1 at 8:07
Michael Rozenberg
97.5k1589188
asked Jan 1 at 6:11
Anu RadhaAnu Radha
1359
In a triangle $ABC$, let $D$ be a point on the segment $BC$ such that $AB + BD = AC + CD$. Suppose that the points $B,C$ and the centroids of triangles $ABD$ and $ACD$ lie on a circle. Prove that $AB = AC$.
I began this way: Denote the centroid of triangle $ABD$ by $G_1$ and the centroid of triangle $ACD$ by $G_2$. Then by using $Basic$ $Proportionality$ $Theorem$, I got that $BG_1G_2C$ is an isosceles trapezium. After that I am stuck. Please help.
geometry triangle
geometry triangle
edited Jan 1 at 8:07
Michael Rozenberg
97.5k1589188
asked Jan 1 at 6:11
Anu RadhaAnu Radha
1359
edited Jan 1 at 8:07
Michael Rozenberg
97.5k1589188
asked Jan 1 at 6:11
Anu RadhaAnu Radha
1359
edited Jan 1 at 8:07
Michael Rozenberg
97.5k1589188
edited Jan 1 at 8:07
Michael Rozenberg
97.5k1589188
edited Jan 1 at 8:07
Michael Rozenberg
97.5k1589188
97.5k1589188
asked Jan 1 at 6:11
Anu RadhaAnu Radha
1359
asked Jan 1 at 6:11
Anu RadhaAnu Radha
1359
asked Jan 1 at 6:11
Anu RadhaAnu Radha
1359
1359
Try to prove that BD=CD
– Moti
Jan 1 at 7:08
Yup, I tried that, too, @Moti. But I feel I am missing something which I could gather from the given information (or by any other manipulation)
– Anu Radha
Jan 1 at 7:39
add a comment |
Try to prove that BD=CD
– Moti
Jan 1 at 7:08
Yup, I tried that, too, @Moti. But I feel I am missing something which I could gather from the given information (or by any other manipulation)
– Anu Radha
Jan 1 at 7:39
Try to prove that BD=CD
– Moti
Jan 1 at 7:08
Try to prove that BD=CD
– Moti
Jan 1 at 7:08
Yup, I tried that, too, @Moti. But I feel I am missing something which I could gather from the given information (or by any other manipulation)
– Anu Radha
Jan 1 at 7:39
Yup, I tried that, too, @Moti. But I feel I am missing something which I could gather from the given information (or by any other manipulation)
– Anu Radha
Jan 1 at 7:39
add a comment |
1 Answer
1
active
oldest
votes
Now, in the standard notation $$c+BD=b+a-BD,$$ which gives
$$BD=frac{a+b-c}{2}$$ and $$CD=frac{a+c-b}{2}.$$
Thus, $$CG_2=frac{1}{3}sqrt{2b^2+2left(frac{a+c-b}{2}right)^2-AD^2}$$ and
$$BG_1=frac{1}{3}sqrt{2c^2+2left(frac{a+b-c}{2}right)^2-AD^2},$$ which gives
$$2b^2+2left(frac{a+c-b}{2}right)^2-AD^2=2c^2+2left(frac{a+b-c}{2}right)^2-AD^2$$ or
$$(b-c)(b+c-a)=0$$ or $$b=c.$$
answered Jan 1 at 8:04
Michael RozenbergMichael Rozenberg
97.5k1589188
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Now, in the standard notation $$c+BD=b+a-BD,$$ which gives
$$BD=frac{a+b-c}{2}$$ and $$CD=frac{a+c-b}{2}.$$
Thus, $$CG_2=frac{1}{3}sqrt{2b^2+2left(frac{a+c-b}{2}right)^2-AD^2}$$ and
$$BG_1=frac{1}{3}sqrt{2c^2+2left(frac{a+b-c}{2}right)^2-AD^2},$$ which gives
$$2b^2+2left(frac{a+c-b}{2}right)^2-AD^2=2c^2+2left(frac{a+b-c}{2}right)^2-AD^2$$ or
$$(b-c)(b+c-a)=0$$ or $$b=c.$$
answered Jan 1 at 8:04
Michael RozenbergMichael Rozenberg
97.5k1589188
add a comment |
Now, in the standard notation $$c+BD=b+a-BD,$$ which gives
$$BD=frac{a+b-c}{2}$$ and $$CD=frac{a+c-b}{2}.$$
Thus, $$CG_2=frac{1}{3}sqrt{2b^2+2left(frac{a+c-b}{2}right)^2-AD^2}$$ and
$$BG_1=frac{1}{3}sqrt{2c^2+2left(frac{a+b-c}{2}right)^2-AD^2},$$ which gives
$$2b^2+2left(frac{a+c-b}{2}right)^2-AD^2=2c^2+2left(frac{a+b-c}{2}right)^2-AD^2$$ or
$$(b-c)(b+c-a)=0$$ or $$b=c.$$
answered Jan 1 at 8:04
Michael RozenbergMichael Rozenberg
97.5k1589188
add a comment |
Now, in the standard notation $$c+BD=b+a-BD,$$ which gives
$$BD=frac{a+b-c}{2}$$ and $$CD=frac{a+c-b}{2}.$$
Thus, $$CG_2=frac{1}{3}sqrt{2b^2+2left(frac{a+c-b}{2}right)^2-AD^2}$$ and
$$BG_1=frac{1}{3}sqrt{2c^2+2left(frac{a+b-c}{2}right)^2-AD^2},$$ which gives
$$2b^2+2left(frac{a+c-b}{2}right)^2-AD^2=2c^2+2left(frac{a+b-c}{2}right)^2-AD^2$$ or
$$(b-c)(b+c-a)=0$$ or $$b=c.$$
answered Jan 1 at 8:04
Michael RozenbergMichael Rozenberg
97.5k1589188
Now, in the standard notation $$c+BD=b+a-BD,$$ which gives
$$BD=frac{a+b-c}{2}$$ and $$CD=frac{a+c-b}{2}.$$
Thus, $$CG_2=frac{1}{3}sqrt{2b^2+2left(frac{a+c-b}{2}right)^2-AD^2}$$ and
$$BG_1=frac{1}{3}sqrt{2c^2+2left(frac{a+b-c}{2}right)^2-AD^2},$$ which gives
$$2b^2+2left(frac{a+c-b}{2}right)^2-AD^2=2c^2+2left(frac{a+b-c}{2}right)^2-AD^2$$ or
$$(b-c)(b+c-a)=0$$ or $$b=c.$$
answered Jan 1 at 8:04
Michael RozenbergMichael Rozenberg
97.5k1589188
answered Jan 1 at 8:04
Michael RozenbergMichael Rozenberg
97.5k1589188
answered Jan 1 at 8:04
Michael RozenbergMichael Rozenberg
97.5k1589188
answered Jan 1 at 8:04
Michael RozenbergMichael Rozenberg
97.5k1589188
97.5k1589188
add a comment |
add a comment |
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Try to prove that BD=CD
– Moti
Jan 1 at 7:08
Yup, I tried that, too, @Moti. But I feel I am missing something which I could gather from the given information (or by any other manipulation)
– Anu Radha
Jan 1 at 7:39