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3 years ago

Geometry - Triangle d) In the given figure, BE = EC and CE is the bisector of ZACB. Prove that ZBEC = ZACD. E B D С

Mathematics

1 answer:

Llana [10]3 years ago

8 0

Answer:

Let m∠BCE = x

Then m∠ACE = x as well since CE is bisecting ∠ACB.

m∠ACD + x + x = 180° ⇒
m∠ACD = 180° - 2x

Consider ΔBEC

Since BE = EC, the opposite angles are congruent:

∠BCE ≅ ∠CBE

Then:

m∠CBE = m∠BCE = x

Find the angle BEC:

m∠BEC = 180° - (x + x) = 180° - 2x

Comparing the above we see that:

m∠BEC = m∠ACD
proved

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Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=Simplify%3A%20%5Cfrac%7B%205%C3%97%2825%29%5E%7Bn%2B1%7D%20-%2025%20%C3%97%20%285%29%5E%7B2n%7

Katen [24]

$\green{\large\underline{\sf{Solution-}}}$

Given expression is 

$\rm :\longmapsto\:\dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} }$

can be rewritten as

$\rm \: = \: \dfrac{5 \times { {(5}^{2} )}^{n + 1} - {5}^{2} \times {5}^{2n} }{5 \times {5}^{2n + 3} - {( {5}^{2} )}^{n + 1} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {( {x}^{m} )}^{n} \: = \: {x}^{mn}}}} \\$

And

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }}} \\$

So, using this identity, we

$\rm \: = \: \dfrac{5 \times {5}^{2n + 2} - {5}^{2n + 2} }{{5}^{2n + 3 + 1} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 4} - {5}^{2n + 2} }$

can be further rewritten as

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 2 + 2} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{ {5}^{2n + 2} (5 - 1)}{ {5}^{2n + 2} ( {5}^{2} - 1)}$

$\rm \: = \: \dfrac{4}{25 - 1}$

$\rm \: = \: \dfrac{4}{24}$

$\rm \: = \: \dfrac{1}{6}$

Hence, 

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} } = \frac{1}{6} }}$

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3 years ago

Geometry - Triangle d) In the given figure, BE = EC and CE is the bisector of ZACB. Prove that ZBEC = ZACD. E B D С​

Geometry - Triangle d) In the given figure, BE = EC and CE is the bisector of ZACB. Prove that ZBEC = ZACD. E B D С