Question

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.​

Answer 1

Given

• Square ABCD with Diagonal BD
• ∆BCE which is described on baseBC
• ∆BDF which is descried on base BD
• Both ∆ BCE and ∆ BDF equilateral

To Prove

• $\: \bf \dfrac{Area. ~∆ BCE }{Area.~ ∆ FDB }= \dfrac{1}{2}$

Proof:

$~~~~~~ \gray{ \footnotesize \textsf{Both ∆BCE and ∆BDF equilateral \: \: \: \: \: - - - - (given)}}$

Now In ∆BCE and ∆BDF ,

$~~~~~~~~~ \footnotesize \sf \dfrac{ DF }{ CE } = \dfrac{ FB }{ EB } = \dfrac{ DB }{ CB }$

So, By SSS Similarity ∆BCE ∼ ∆BDF

We know that in similar triangles,

• Ratio of area of triangle is equal to ratio of square of corresponding sides

$\footnotesize \sf\dfrac{ Area. ~ ∆ FBD}{Area.~∆ BCE}={\bigg( \dfrac{DB}{BC}\bigg) }^{2} \: \: \: \: \: - - - - (1)$

_____________________

We know that

• If ABCD is a square then the diagonal is √2 times the Side .

$\footnotesize \sf DB = \sqrt{2} BC$

_____________________

• Putting the value of DB in Equ. (1)

$~~~~~ \mapsto \footnotesize \sf\dfrac{ Area. ~ ∆ FBD}{Area.~∆ BCE}={\bigg( \dfrac{ \sqrt{2} BC }{BC}\bigg) }^{2}$

$~~~~~ \mapsto \footnotesize \sf\dfrac{ Area. ~ ∆ FBD}{Area.~∆ BCE}={\bigg( \dfrac{ \sqrt{2} \: \cancel{BC} }{ \cancel{BC}}\bigg) }^{2}$

$~~~~~ \mapsto \footnotesize \sf\dfrac{ Area. ~ ∆ FBD}{Area.~∆ BCE}={\bigg( \sqrt{2} \times \sqrt{2} \bigg) }$

$~~~~~ \mapsto \footnotesize \sf\dfrac{ Area. ~ ∆ FBD}{Area.~∆ BCE}=2$

${ \small\bf\dfrac{ Area.~∆ BCE }{Area. ~ ∆ FBD}= \dfrac{1}{2}} _{_{_{_{_{_{ \normalsize \pink{Hence~~ proved}}}}}}}$

Answer 2

Answer:

hey here is your proof in above pics

so pls mark it as brainliest