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
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Absolutely! Let's break down this geometry problem step-by-step.
**Understanding the Problem**
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.
**Diagram**
It's crucial to visualize this problem. Here's a description of the diagram:
1. Start with triangle DEF.
2. Extend DE to G and DF to H.
3. EF = EG = FH.
4. EH and FG intersect at K.
**Solution**
1. **Isosceles Triangles:**
* Since EG = EF, triangle EFG is isosceles. Therefore, ∠EFG = ∠EGF.
* Since FH = EF, triangle EFH is isosceles. Therefore, ∠FEH = ∠FHE.
2. **Angle Sum of Triangle DEF:**
* In triangle DEF, ∠DEF + ∠DFE + ∠EDF = 180°.
* Let ∠DEF = a and ∠DFE = b.
* Then, a + b + 2x = 180°.
* Therefore, a + b = 180° - 2x.
3. **Angles in Isosceles Triangles:**
* In triangle EFG, let ∠EFG = ∠EGF = y.
* In triangle EFH, let ∠FEH = ∠FHE = z.
* The angles on a straight line add to 180 degrees.
* ∠DEG = 180 - a, and ∠DFH = 180 -b.
* In triangle EFG, 2y + 180-a = 180, so 2y = a, and y = a/2.
* In triangle EFH, 2z + 180-b = 180, so 2z = b, and z = b/2.
4. **Angles in Triangle EKF**
* ∠EKF = 180 - (z + y)
* ∠EKF = 180 - (b/2 + a/2)
* ∠EKF = 180 - (a+b)/2
5. **Substitution**
* We know a + b = 180 - 2x
* Substitute this into the equation for ∠EKF:
* ∠EKF = 180 - (180 - 2x)/2
* ∠EKF = 180 - (90 - x)
* ∠EKF = 180 - 90 + x
* ∠EKF = 90 + x
6. **Vertical Angles**
* ∠EKG and ∠EKF are supplementary angles, so ∠EKG + ∠EKF = 180.
* ∠EKG = 180 - ∠EKF
* ∠EKG = 180 - (90 + x)
* ∠EKG = 180 - 90 - x
* ∠EKG = 90 - x.
**Therefore, we have shown that angle EKG = 90 - x degrees.**
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