Question

there is a triangle abd with ab=7, ad=5, and a triangle cbd with cb=7, cd=4. then there is a multiple - choice question: compare m∠abd and m∠cbd, and there is an option cannot be determined. then there is an isosceles triangle efg with ef=fg, ∠f = 116°, and an isosceles triangle xyz with xy=yz, ∠y = 114°. then there is a multiple - choice question: compare eg and xz, and there is an option cannot be determined. there are also some incomplete triangle figures at the bottom.

Explanation:

Step1: Analyze $\boldsymbol{\angle ABD}$ and $\boldsymbol{\angle CBD}$

In $\triangle ABD$ and $\triangle CBD$, $AB = CB = 7$, $BD$ is common. $AD = 5$, $CD = 4$. By the Hinge Theorem (SAS Inequality Theorem), if two sides of one triangle are congruent to two sides of another triangle, but the third side is longer, the included angle is larger. Here, $AD > CD$ ($5 > 4$), so $m\angle ABD > m\angle CBD$.

Step2: Analyze $\boldsymbol{EG}$ and $\boldsymbol{XZ}$

In $\triangle EFG$, $EF = FG$ (marked congruent), so it's isosceles with base angles equal. The vertex angle $\angle F = 116^\circ$, so base angles $\angle E = \angle G=\frac{180 - 116}{2}=32^\circ$. In $\triangle XYZ$, $XY = YZ$ (marked congruent), vertex angle $\angle Y = 114^\circ$, so base angles $\angle X = \angle Z=\frac{180 - 114}{2}=33^\circ$. Now, in $\triangle EFG$ and $\triangle XYZ$, two sides are congruent (marked), but the included angles: $\angle F = 116^\circ$, $\angle Y = 114^\circ$. By the Hinge Theorem, since $\angle F > \angle Y$, the side opposite $\angle F$ (which is $EG$) and the side opposite $\angle Y$ (which is $XZ$): wait, no—wait, in $\triangle EFG$, sides $EF = FG$, included angle $\angle F$; in $\triangle XYZ$, sides $XY = YZ$, included angle $\angle Y$. Wait, actually, the sides: $EF = XY$ (marked), $FG = YZ$ (marked). So by SAS, but angles: $\angle F = 116^\circ$, $\angle Y = 114^\circ$. So the side opposite the larger angle: in $\triangle EFG$, $EG$ is opposite $\angle F$; in $\triangle XYZ$, $XZ$ is opposite $\angle Y$. Wait, no—wait, in $\triangle EFG$, sides $EF$ and $FG$ with included angle $\angle F$; in $\triangle XYZ$, sides $XY$ and $YZ$ with included angle $\angle Y$. So the third sides: $EG$ (opposite $\angle F$) and $XZ$ (opposite $\angle Y$). Since $\angle F > \angle Y$, $EG > XZ$? Wait, no, wait: Hinge Theorem says if two sides are congruent, and included angle is larger, then the third side is larger. So $EF = XY$, $FG = YZ$, $\angle F > \angle Y$, so $EG > XZ$. Wait, but let's recast: $\triangle EFG$: $EF = FG$, $\angle F = 116^\circ$, so $EG^2 = EF^2 + FG^2 - 2\cdot EF\cdot FG\cdot \cos(116^\circ)$. $\triangle XYZ$: $XY = YZ$, $\angle Y = 114^\circ$, so $XZ^2 = XY^2 + YZ^2 - 2\cdot XY\cdot YZ\cdot \cos(114^\circ)$. Since $\cos(116^\circ) < \cos(114^\circ)$ (because $116^\circ > 114^\circ$, and cosine is decreasing in $[0, 180]$), so $-2\cdot EF\cdot FG\cdot \cos(116^\circ) > -2\cdot XY\cdot YZ\cdot \cos(114^\circ)$ (since $EF = XY$, $FG = YZ$), so $EG^2 > XZ^2$, so $EG > XZ$.

Answer:

For $m\angle ABD$ and $m\angle CBD$: $m\angle ABD > m\angle CBD$
For $EG$ and $XZ$: $EG > XZ$