Question

there are three geometric figures with related angle and side comparison questions. first figure: quadrilateral with sides ab=12, ad=8, bc=8, cd=16, diagonal bd. question: compare m∠adb and m∠cbd (options: <, >, =, or cannot be determined). second figure: triangle efg with ef=8, fg=7, ∠f=112°; triangle xyz with yz=8, xz=7, ∠z=104°. question: compare eg and xy (options: <, >, =, or cannot be determined). third figure: triangle jkl with jl=11, two marked equal sides; triangle pqr with pq=14, two marked equal sides. question: compare m∠k and m∠r (options: <, >, =, or cannot be determined).

Explanation:

First Comparison (∠ADB and ∠CBD)

Step1: Analyze Triangle ABD and CBD

In quadrilateral \(ABCD\), \(AB = 12\), \(AD = 8\), \(BC = 8\), \(CD = 16\). Notice that \( \triangle ABD \) and \( \triangle CBD \): \(AD = BC = 8\), \(AB = 12\), \(BD\) is common. Also, \( \angle ADB \) and \( \angle CBD \) are alternate - interior angles if \(AD\parallel BC\)? Wait, let's check the sides. \( \frac{AD}{BC}=\frac{8}{8} = 1\), \( \frac{AB}{CD}=\frac{12}{16}=\frac{3}{4}\)? Wait, no, actually, in \( \triangle ABD \) and \( \triangle CDB \), \(AD = BC = 8\), \(BD = BD\), and \(AB = 12\), \(CD = 16\)? Wait, no, maybe we can use the converse of alternate - interior angles. Since \(AD = 8\), \(BC = 8\), \(AB = 12\), \(AD\) and \(BC\) are both horizontal (from the diagram). So \(AD\parallel BC\), and \(BD\) is a transversal. So \( \angle ADB\) and \( \angle CBD\) are alternate - interior angles, so they are equal. So \(m\angle ADB=m\angle CBD\).

Second Comparison (EG and XY)

Step1: Analyze Triangles EFG and XYZ

In \( \triangle EFG \), \(EF = 8\), \(FG = 7\), \( \angle F=112^{\circ}\). In \( \triangle XYZ \), \(XZ = 7\), \(YZ = 8\), \( \angle Z = 104^{\circ}\). We can use the Law of Cosines. For \( \triangle EFG\), \(EG^{2}=EF^{2}+FG^{2}-2\cdot EF\cdot FG\cdot\cos\angle F\). For \( \triangle XYZ\), \(XY^{2}=XZ^{2}+YZ^{2}-2\cdot XZ\cdot YZ\cdot\cos\angle Z\). Since \(EF = YZ = 8\), \(FG = XZ = 7\), and \( \cos112^{\circ}<\cos104^{\circ}\) (because \(112^{\circ}>104^{\circ}\) and cosine is decreasing in \(0^{\circ}\) to \(180^{\circ}\)). So \( - 2\cdot EF\cdot FG\cdot\cos\angle F>-2\cdot XZ\cdot YZ\cdot\cos\angle Z\) (because \(\cos112^{\circ}\) is more negative). Then \(EG^{2}=8^{2}+7^{2}-2\times8\times7\times\cos112^{\circ}\), \(XY^{2}=7^{2}+8^{2}-2\times7\times8\times\cos104^{\circ}\). Since \(\cos112^{\circ}<\cos104^{\circ}\), \( - 2\times8\times7\times\cos112^{\circ}>-2\times7\times8\times\cos104^{\circ}\), so \(EG^{2}>XY^{2}\), so \(EG > XY\).

Third Comparison (∠K and ∠R)

Step1: Analyze Triangles JKL and PQR

In \( \triangle JKL \), two sides are marked as equal (isosceles triangle), with one side \(JL = 11\). In \( \triangle PQR \), two sides are marked as equal (isosceles triangle), with one side \(PQ = 14\). Let's denote the equal sides in \( \triangle JKL\) as \(a\) and \(b\) (with \(JL = 11\) as the non - equal side), and in \( \triangle PQR\) as \(c\) and \(d\) (with \(PQ = 14\) as the non - equal side). By the Law of Cosines, for an isosceles triangle with equal sides \(s\) and base \(b\), the vertex angle \(\theta\) satisfies \(\cos\theta=\frac{2s^{2}-b^{2}}{2s^{2}}\). In \( \triangle JKL\), let the equal sides be \(s_1\), base \(b_1 = 11\). In \( \triangle PQR\), equal sides be \(s_2\), base \(b_2 = 14\). Since the equal sides in both triangles seem to be marked similarly (the number of tick marks), assume \(s_1=s_2\). Then, since \(b_1 = 11\cos\angle R\) (because \(2s_1^{2}-11^{2}>2s_1^{2}-14^{2}\)). Since cosine is decreasing in \(0^{\circ}\) to \(180^{\circ}\), if \(\cos\angle K>\cos\angle R\), then \(m\angle K

Answer:

\( m\angle ADB = m\angle CBD \)