Question

In the given figure AB and ED are perpendicular to BD. If AB = ED and BF = DC.
Show that AC = EF.

Answer 1

Answer:

PROVED

Follow the steps ...

Step-by-step explanation:

Given: AB = ED & BF = DC

∴ BC = BF + FC ....... (i)

∴ DF = DC + FC ....... (ii)

Since, BF = DC

Let BC = DC = x

∴ In equation (i):-

BC = x + FC ......... (iii)

∴ In equation (iI):-

DF = x + FC ............. (iv)

Combining equations (iii) and (iv) we get:-

BC = x + FC = DF

∴ BC = DF

Since AB = ED

∴ Let AB  = ED = y

NOW,

In ΔABC,

AC² = AB² + BC²

⇒ AC² = y² + x² ......... (v)

In ΔEDF,

EF² = ED² + FD²

⇒EF² = y² + x² ................. (vi)

From equations (v) and (vi), we get,

AC² = x² + y² = EF²

⇒AC² = EF²

⇒$\sqrt{AC^{2} }$ = $\sqrt{EF^{2} }$    {Square rooting on both sides}

                                           HENCE PROVED