Question

100 POINTS!! OMG YOU NEED TO SOLVE THIS!

Answer 1

Answer:

The correct answer is: $\Delta BDE \cong \Delta BFK$ by rule ASA rule of congruence.

Step-by-step explanation:

First let us prove $\Delta BDE \cong \Delta BFK$ by rule ASA (rule of congruence).

Congruent side:

$\overline{BD} \cong \overline{BF}$ (Given)

Congruent angles:

1. By definition of perpendicular,

$\angle{BFK} = 90 \textdegree \ (Since \ \overline{FK} \ is \ perpendicular \ to \ \overline{AB} \ (\overline{FK} \perp \overline{AB} =Given))$

Also,

$\angle{BDE} = 90 \textdegree \ (Given)$

Therefore,

$\angle{BDE} \cong \angle{BFK}$

or you can say,

$\angle{D} \cong \angle{F}$

2. Common angle between $\Delta BDE \ and \ \Delta BFK$ is $\angle{B}$

In a nutshell, in $\Delta BDE$,

$\angle{B}$ (Angle)

$\overline{BD}$ (Side)

$\angle{BDE}$ (Angle)

are congruent to the following angle, side and angle of $\Delta BFK$:

$\angle{B}$ (Angle)

$\overline{BF}$ (Side)

$\angle{BFK}$ (Angle)

Therefore, by ASA rule of congruence, we can say $\Delta BDE \cong \Delta BFK$.

Since both triangles are congruent, the sides $\overline{ED}$ and $\overline{FK}$ are also congruent.

Answer 2

Answer:

This would be 90 degeree angle

Step-by-step explanation: