Question

given: $overline{af}perpoverline{bd}$, $overline{ce}perpoverline{bd}$, $ab = cd$, $be = fd$. prove: $\triangle abfcong\triangle cde$. 1. $overline{af}perpoverline{bd}$, $overline{ce}perpoverline{bd}$ 1. given 2. click here to insert 2. perpendicular lines intersect to form right angles. 3. click here to insert 3. definition of a right triangle 4. $be = bf + fe$ and $fd = de + fe$ 4. click here to insert

Explanation:

Step1: Define right - angles

Since $\overline{AF}\perp\overline{BD}$ and $\overline{CE}\perp\overline{BD}$, by the property of perpendicular lines, $\angle AFB=\angle CED = 90^{\circ}$.

Step2: Identify right - triangles

By the definition of a right - triangle (a triangle with one right angle), $\triangle ABF$ and $\triangle CDE$ are right - triangles.

Step3: Apply segment addition

The segment addition postulate states that for three collinear points, the length of the whole segment is the sum of the lengths of its sub - segments. So $BE = BF + FE$ and $FD = DE+FE$.

Step4: Prove equal segments

Given $BE = FD$, substituting the segment addition expressions $BF + FE=DE + FE$. Subtracting $FE$ from both sides gives $BF=DE$.

Step5: Prove triangle congruence

In right - triangles $\triangle ABF$ and $\triangle CDE$, we have the hypotenuse $AB = CD$ (given) and a leg $BF = DE$ (proven). By the Hypotenuse - Leg (HL) congruence theorem for right - triangles, $\triangle ABF\cong\triangle CDE$.

Answer:

1. $\angle AFB=\angle CED = 90^{\circ}$
2. $\triangle ABF$ and $\triangle CDE$ are right - triangles
3. Segment addition postulate
4. Since $BE = FD$, then $BF+FE=DE + FE$, so $BF=DE$
5. In right - triangles $\triangle ABF$ and $\triangle CDE$, $AB = CD$ (given) and $BF=DE$ (proven above), so $\triangle ABF\cong\triangle CDE$ (HL congruence theorem)