Answer:
∠B ≅ ∠F ⇒ proved down
Step-by-step explanation:
<em>In the </em><em>two right triangles</em><em>, if the </em><em>hypotenuse and leg</em><em> of the </em><em>1st right Δ ≅</em><em> the </em><em>hypotenuse and leg</em><em> of the </em><em>2nd right Δ</em><em>, then the </em><em>two triangles are congruent</em>
Let us use this fact to solve the question
→ In Δs BCD and FED
∵ ∠C and ∠E are right angles
∴ Δs BCD and FED are right triangles ⇒ (1)
∵ D is the mid-point of CE
→ That means point D divides CE into 2 equal parts CD and ED
∴ CD = ED ⇒ (2) legs
∵ BD and DF are the opposite sides to the right angles
∴ BD and DF are the hypotenuses of the triangles
∵ BD ≅ FD ⇒ (3) hypotenuses
→ From (1), (2), (3), and the fact above
∴ Δ BCD ≅ ΔFED ⇒ by HL postulate of congruency
→ As a result of congruency
∴ BC ≅ FE
∴ ∠BDC ≅ ∠FDE
∴ ∠B ≅ ∠F ⇒ proved
309/17 equals about 18.176 but it’s a rounded answer/decimal.
Check the picture below.
so, bearing in mind that vertical angles are equal, thus ∡AED = ∡BEC, and we know that ∡BEC = 52 = ∡AED, so those two angles added up are 104°.
now, a circle has a total of 360°, if we take away those 104° from the red angles, what's leftover is just 256°, well, the green angles there are also vertical angles and thus twins, so each one will take half of 256°.
∡BEA = 128° = ∡CED.
Answer:
all are functions except the graph because if u use a ruler it touches the lines more than once at one time
Step-by-step explanation:
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