Given

Prove:

lwfrgin
2021-08-09
Answered

Write a proof for the following:

Given$\stackrel{\u2015}{AB}\stackrel{\sim}{=}\stackrel{\u2015}{DE}$ and C is the midpoint of both $\stackrel{\u2015}{AB}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}\stackrel{\u2015}{BE}$ .

Prove:$\mathrm{\u25b3}ABC\stackrel{\sim}{=}\mathrm{\u25b3}DEC$ .

Given

Prove:

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curwyrm

Answered 2021-08-10
Author has **87** answers

Step 1

$AB\stackrel{\sim}{=}DE$ and C is the mid point of both AD and BE.

Now

AC=DC...(1) ($\because$ C is mid point of AD)

CE=BC...(2) ($\because$ C is the mid point of BE)

$\mathrm{\angle}ACB=\mathrm{\angle}BCE$ ...(3) ($\because$ vertically opposite $\mathrm{\angle}s$ )

Also$AB\stackrel{\sim}{=}DE$ ...(4) (Given)

Step 2

Therefore from (1), (2) and (4)

$\mathrm{\u25b3}ABC\stackrel{\sim}{=}\mathrm{\u25b3}DEC$ (by SSS similarity)

Also from (1), (2) and (4)

$\mathrm{\u25b3}ABC\stackrel{\sim}{=}\mathrm{\u25b3}DEC$ (by SAS similarity)

Now

AC=DC...(1) (

CE=BC...(2) (

Also

Step 2

Therefore from (1), (2) and (4)

Also from (1), (2) and (4)

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