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oksano4ka [1.4K]
3 years ago
11

Copy the problems onto your paper, mark the given and prove the statements asked. Given: marked Prove: ΔABC ≅ ΔDBC, ΔEHF ≅ ΔGHF

Mathematics
1 answer:
hjlf3 years ago
7 0

The proof is given below. Please go through it.

Step-by-step explanation:

To solve Δ ABC ≅ Δ DBC

From Δ ABC and Δ DBC

AB = BD (given)

AC = CD (given)

BC is common side

By SSS condition Δ ABC ≅ Δ DBC ( proved)

To solve Δ EHF ≅ Δ GHF

Δ EHF and Δ GHF

EH = HG ( given)

∠ EFH = ∠ GFH ( each angle is 90°)

HF is common side

By RHS condition

Δ EHF ≅ Δ GHF

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d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(0-2)^2 + (-1-1)^2}\\\\d = \sqrt{(-2)^2 + (-2)^2}\\\\d = \sqrt{4 + 4}\\\\d = \sqrt{8}\\\\d \approx 2.828\\\\

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Through very similar steps, you should find that the other side lengths of BC, CD and AD are all the same.

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-----------------------------

Let's find the slope of line AB

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{1 - (-1)}{2 - 0}\\\\m = \frac{1 + 1}{2 - 0}\\\\m = \frac{2}{2}\\\\m = 1\\\\

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Therefore, this figure is a rectangle. Any rectangle has all four angles of 90 degrees.

-----------------------------

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Therefore, quadrilateral ABCD is a square.

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