Math / Geometry

Question1. Given: parallelogram FLSH, diagonal FGAS, $\overline{L G} \perp \overline{F S}, \overline{H A} \perp \overline{F S}$ Prove: $\triangle L G S \equiv \triangle H A F$

Studdy Solution

STEP 1

1. Parallelogram FLSH has a diagonal FGAS.
2. $\overline{L G} \perp \overline{F S}$ and $\overline{H A} \perp \overline{F S}$ .
3. Triangles LGS and HAF are right triangles because of the perpendicular lines.

STEP 2

1. Identify the right triangles and their components.
2. Use the properties of parallelograms to establish congruence.
3. Apply the criteria for triangle congruence.

STEP 3

Identify the right triangles and their components:
- $\triangle LGS$ and $\triangle HAF$ are right triangles because $\overline{L G} \perp \overline{F S}$ and $\overline{H A} \perp \overline{F S}$ .

STEP 4

Use the properties of parallelograms to establish congruence:
- In parallelogram FLSH, opposite sides are equal: $\overline{FL} = \overline{SH}$ and $\overline{FS} = \overline{LH}$ . - Since $\overline{L G} \perp \overline{F S}$ and $\overline{H A} \perp \overline{F S}$ , $\angle LGS = \angle HAF = 90^\circ$ .

STEP 5

Apply the criteria for triangle congruence:
- By the Hypotenuse-Leg (HL) theorem for right triangles: - Hypotenuses $\overline{LS}$ and $\overline{HF}$ are equal because they are opposite sides of parallelogram FLSH. - Legs $\overline{LG}$ and $\overline{HA}$ are perpendicular to the same line $\overline{FS}$ .
Thus, $\triangle LGS \equiv \triangle HAF$ by the HL theorem.
The triangles $\triangle LGS$ and $\triangle HAF$ are congruent.

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