Question

A(7, 1), B(3, 5) are the end points of a line segment. C is a point on AB such that CD is tho perpondicular bisector of AB and E, F are midpoints of AC and BC, Then C=?,EF=?​

Answer 1

Answer:

8

Step-by-step explanation:

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Answer 2

C is equidistant from points A and B

CA=CB and D is equidistant from points A and B

⟹DA=DB

We have to prove that CD⊥AB

AD=BD and ∠CPA=∠CPB=90

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In △CAD and △CBD

AC=BC ....... (1)

AD=BD ........ (2)

⟹CD=CD ......... (common)

∴△CAD≅△CBD ..... (SSS congruence rule)

Hence ∠ACD=∠BCD ........ (C.P.C.T) (3)

In △CAP≅△CBP ........ (SAS congruence rule)

∴AP=BP(C.P.C.T)

⟹∠APC=∠BPC ........ (C.P.C.T)

Since AB is a line segment

∠APC+∠BPC=180

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⟹2∠APC=180

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∴∠APC=90

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∠APC=∠BPC=90

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Hence, AC=BC and ∠APC=∠BPC=90

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∴CD is ⊥ bisector of AB