Question

prove that the area of eq triangle described on one side of square is equal to half the area of eq triangle described on one of it's diagonals​

Answer 1

★★$\huge\bold\red{{Question:-}}$

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

★★$\huge\bold\orange{{Given:-}}$

• Square ABCD with diagonal BD

• ∆ BCE which is described on base BC

• ∆BDF which is described on base BD

• Both ∆ BCE and ∆ BDF are equilateral

★★$\huge\bold\purple{{To\:Prove:-}}$

$\sf\:⇢\dfrac{ar\:{\triangle\:BCE}}{ar\:{\triangle\:FDB}}=\dfrac{1}{2}$

⠀ ⠀ ⠀⠀

★★$\huge\bold\green{{Proof:-}}$

Both ∆ BCE and ∆ BDF are equilateral

In ∆ BDF and ∆ BCE

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

hence, by SSS similarity

∆ FDB ≅ ∆ BCE

we know that in similar triangle,

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

$\sf\:\dfrac{ar\:{\triangle\:FBD}}{ar\:{\triangle\:BCE}}=\dfrac{DB}{BC}^2$

But,$\sf\:DB={ \sqrt{2}}BC$ as DB is the diagonal of square ABCD

hence,

$\sf\:\dfrac{ar\:{\triangle\:FBD}}{ar\:{\triangle\:BCE}}$=(√2BC/BC)²

$\sf\:\dfrac{ar\:{\triangle\:FBD}}{ar\:{\triangle\:BCE}}=({ \sqrt{2}})^2$

$\sf\:\dfrac{ar\:{\triangle\:FBD}}{ar\:{\triangle\:BCE}}=2$

$\sf\:\bold\pink{{\dfrac{ar\:{\triangle\:FBD}}{ar\:{\triangle\:BCE}}=\dfrac{1}{2}}}$✔️✔️

hence proved.