Question

which of the following statements is true? select all that apply.

(image of two triangles: triangle jkl with sides jk=9, jl=8, kl=9; triangle mno with sides mn=7, mo=9, no=9)

• ∠j ≅ ∠l
• the circumcenter for each triangle is located inside the triangle.
• ∠j ≅ ∠m
• m∠k > m∠o

Explanation:

Step1: Analyze $\angle J \cong \angle L$

In $\triangle JKL$, $JK = LK = 9$, so it's isosceles with $\angle J \cong \angle L$. Correct.

Step2: Analyze circumcenter location

$\triangle JKL$: isosceles (acute? $8,9,9$: by Pythagoras, $8^2 + 8^2 > 9^2$? Wait, sides $8,9,9$. Angles: base angles at $J,L$; vertex at $K$. All angles acute? Circumcenter inside. $\triangle MNO$: sides $7,9,9$: also isosceles, acute. Circumcenter inside? Wait, no—wait, the statement says "for each triangle". But wait, maybe I made a mistake? Wait, no, both are acute isosceles, so circumcenter inside. But the option was unchecked. Wait, no, the original problem's checked options: wait, the user's image shows some checked. Wait, no, let's re-express. Wait, the first step: $\angle J \cong \angle L$: correct (isosceles $\triangle JKL$, $JK=LK$).

Step3: Analyze $\angle J \cong \angle M$

$\triangle JKL$: sides $8,9,9$; $\triangle MNO$: sides $7,9,9$. Let's find angles. In $\triangle JKL$, angle at $J$: opposite $LK=9$; in $\triangle MNO$, angle at $M$: opposite $NO=9$. Wait, $\triangle JKL$: sides $8,9,9$; $\triangle MNO$: $7,9,9$. The angle opposite $9$ in $\triangle JKL$: angle $J$ (opposite $LK=9$); in $\triangle MNO$, angle $M$ (opposite $NO=9$). Wait, but the other sides: $\triangle JKL$ has side $8$ (between $J$ and $L$), $\triangle MNO$ has side $7$ (between $N$ and $M$). Wait, using the Law of Cosines: for $\angle J$ in $\triangle JKL$: $\cos J = \frac{8^2 + 9^2 - 9^2}{2 \cdot 8 \cdot 9} = \frac{64}{144} = \frac{4}{9}$. For $\angle M$ in $\triangle MNO$: $\cos M = \frac{7^2 + 9^2 - 9^2}{2 \cdot 7 \cdot 9} = \frac{49}{126} = \frac{7}{18}$. $\frac{4}{9} \approx 0.444$, $\frac{7}{18} \approx 0.389$. So $\angle J > \angle M$? Wait, no—wait, Law of Cosines: smaller cosine means larger angle. Wait, $\frac{4}{9} > \frac{7}{18}$, so $\angle J < \angle M$? Wait, that contradicts. Wait, maybe I mixed up. Wait, in $\triangle JKL$, sides: $JK=9$, $JL=8$, $LK=9$. So angle at $J$: between $JK=9$ and $JL=8$, opposite $LK=9$. In $\triangle MNO$, sides: $MN=7$, $MO=9$, $NO=9$. Angle at $M$: between $MN=7$ and $MO=9$, opposite $NO=9$. So using Law of Cosines: $\cos J = \frac{JK^2 + JL^2 - LK^2}{2 \cdot JK \cdot JL} = \frac{9^2 + 8^2 - 9^2}{2 \cdot 9 \cdot 8} = \frac{64}{144} = \frac{4}{9}$. $\cos M = \frac{MN^2 + MO^2 - NO^2}{2 \cdot MN \cdot MO} = \frac{7^2 + 9^2 - 9^2}{2 \cdot 7 \cdot 9} = \frac{49}{126} = \frac{7}{18} \approx 0.388$, $\frac{4}{9} \approx 0.444$. Since $\cos J > \cos M$, $\angle J < \angle M$? So $\angle J \cong \angle M$ is false? But the image shows it checked. Wait, maybe I made a mistake. Wait, no, the problem is to select all true. Let's re-express:
- $\angle J \cong \angle L$: true (isosceles $\triangle JKL$, $JK=LK$).
- Circumcenter: both triangles are acute (since all sides squared: $8^2=64$, $9^2=81$; $64 + 81 > 81$ (for $\triangle JKL$: $8,9,9$: $8^2 + 9^2 = 64 + 81 = 145 > 81 = 9^2$; so acute. $\triangle MNO$: $7^2 + 9^2 = 49 + 81 = 130 > 81 = 9^2$; acute. So circumcenter inside both. So that statement is true, but in the image, it's unchecked. Wait, maybe the original problem's options: the user's image has some checked. Wait, the fourth option: $m\angle K > m\angle O$. In $\triangle JKL$, angle $K$: opposite $JL=8$; in $\triangle MNO$, angle $O$: opposite $MN=7$. Since $8 > 7$, by Law of Sines, $\frac{8}{\sin K} = \frac{9}{\sin J}$ and $\frac{7}{\sin O} = \frac{9}{\sin M}$. Since $8 > 7$, $\sin K > \sin O$. Since angles $K$ and $O$ are acute (triangles are acute), so $m\angle K > m\angle O$. True.

Wait, the correct true statements:
1. $\angle J \cong \angle L$: true (isosceles $\triangle JKL$).
2. Circumcenter inside each: true (both acute).
3. $\angle J \cong \angle M$: false (as $\angle J < \angle M$).
4. $m\angle K > m\angle O$: true (since $JL=8 > MN=7$, so angle opposite $8$ ($\angle K$) > angle opposite $7$ ($\angle O$)).

But the image shows $\angle J \cong \angle M$ checked, which is wrong. Wait, maybe I messed up. Let's use Law of Cosines for $\angle J$ and $\angle M$:

In $\triangle JKL$:
$JK = 9$, $JL = 8$, $LK = 9$.
$\cos J = \frac{JK^2 + JL^2 - LK^2}{2 \cdot JK \cdot JL} = \frac{81 + 64 - 81}{2 \cdot 9 \cdot 8} = \frac{64}{144} = \frac{4}{9} \approx 0.444$.

In $\triangle MNO$:
$MN = 7$, $MO = 9$, $NO = 9$.
$\cos M = \frac{MN^2 + MO^2 - NO^2}{2 \cdot MN \cdot MO} = \frac{49 + 81 - 81}{2 \cdot 7 \cdot 9} = \frac{49}{126} = \frac{7}{18} \approx 0.389$.

Since $\cos J > \cos M$, $\angle J < \angle M$, so $\angle J \cong \angle M$ is false.

So the true statements are:
- $\angle J \cong \angle L$
- The circumcenter for each triangle is located inside the triangle.
- $m\angle K > m\angle O$

But in the image, $\angle J \cong \angle M$ is checked (wrong), and circumcenter is unchecked (wrong). Maybe the original problem's options: let's re-express the correct ones.

Answer:

The true statements are:

• $\boldsymbol{\angle J \cong \angle L}$ (isosceles $\triangle JKL$, $JK = LK$)
• $\boldsymbol{\text{The circumcenter for each triangle is located inside the triangle}}$ (both triangles are acute, so circumcenter inside)
• $\boldsymbol{m\angle K > m\angle O}$ (since $JL = 8 > MN = 7$, by Law of Sines, $m\angle K > m\angle O$)

(Note: $\angle J \cong \angle M$ is false, as shown by Law of Cosines.)