document.write( "Question 1169822: Please help me to solve this.. DEF is a triangle with angle EDF=2x°. Line DE and line DF are produced to G and H respectively so that side EF=side EG=side FH. Line EH and line FG intersect at K. Show that angle EKG=90-x°. Thank you \n" ); document.write( "
Algebra.Com's Answer #851287 by CPhill(1959)\"\" \"About 
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Absolutely! Let's break down this geometry problem step-by-step.\r
\n" ); document.write( "\n" ); document.write( "**Understanding the Problem**\r
\n" ); document.write( "\n" ); document.write( "We're given a triangle DEF with specific conditions regarding the extensions of its sides and the lengths of certain segments. We need to prove that the angle EKG is equal to 90 - x degrees.\r
\n" ); document.write( "\n" ); document.write( "**Diagram**\r
\n" ); document.write( "\n" ); document.write( "It's crucial to visualize this problem. Here's a description of the diagram:\r
\n" ); document.write( "\n" ); document.write( "1. Start with triangle DEF.
\n" ); document.write( "2. Extend DE to G and DF to H.
\n" ); document.write( "3. EF = EG = FH.
\n" ); document.write( "4. EH and FG intersect at K.\r
\n" ); document.write( "\n" ); document.write( "**Solution**\r
\n" ); document.write( "\n" ); document.write( "1. **Isosceles Triangles:**
\n" ); document.write( " * Since EG = EF, triangle EFG is isosceles. Therefore, ∠EFG = ∠EGF.
\n" ); document.write( " * Since FH = EF, triangle EFH is isosceles. Therefore, ∠FEH = ∠FHE.\r
\n" ); document.write( "\n" ); document.write( "2. **Angle Sum of Triangle DEF:**
\n" ); document.write( " * In triangle DEF, ∠DEF + ∠DFE + ∠EDF = 180°.
\n" ); document.write( " * Let ∠DEF = a and ∠DFE = b.
\n" ); document.write( " * Then, a + b + 2x = 180°.
\n" ); document.write( " * Therefore, a + b = 180° - 2x.\r
\n" ); document.write( "\n" ); document.write( "3. **Angles in Isosceles Triangles:**
\n" ); document.write( " * In triangle EFG, let ∠EFG = ∠EGF = y.
\n" ); document.write( " * In triangle EFH, let ∠FEH = ∠FHE = z.
\n" ); document.write( " * The angles on a straight line add to 180 degrees.
\n" ); document.write( " * ∠DEG = 180 - a, and ∠DFH = 180 -b.
\n" ); document.write( " * In triangle EFG, 2y + 180-a = 180, so 2y = a, and y = a/2.
\n" ); document.write( " * In triangle EFH, 2z + 180-b = 180, so 2z = b, and z = b/2.\r
\n" ); document.write( "\n" ); document.write( "4. **Angles in Triangle EKF**
\n" ); document.write( " * ∠EKF = 180 - (z + y)
\n" ); document.write( " * ∠EKF = 180 - (b/2 + a/2)
\n" ); document.write( " * ∠EKF = 180 - (a+b)/2\r
\n" ); document.write( "\n" ); document.write( "5. **Substitution**
\n" ); document.write( " * We know a + b = 180 - 2x
\n" ); document.write( " * Substitute this into the equation for ∠EKF:
\n" ); document.write( " * ∠EKF = 180 - (180 - 2x)/2
\n" ); document.write( " * ∠EKF = 180 - (90 - x)
\n" ); document.write( " * ∠EKF = 180 - 90 + x
\n" ); document.write( " * ∠EKF = 90 + x\r
\n" ); document.write( "\n" ); document.write( "6. **Vertical Angles**
\n" ); document.write( " * ∠EKG and ∠EKF are supplementary angles, so ∠EKG + ∠EKF = 180.
\n" ); document.write( " * ∠EKG = 180 - ∠EKF
\n" ); document.write( " * ∠EKG = 180 - (90 + x)
\n" ); document.write( " * ∠EKG = 180 - 90 - x
\n" ); document.write( " * ∠EKG = 90 - x.\r
\n" ); document.write( "\n" ); document.write( "**Therefore, we have shown that angle EKG = 90 - x degrees.**
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