Question

given: $overline{af}perpoverline{bd}$, $overline{ce}perpoverline{bd}$, $ab = cd$, $be = fd$. prove: $\triangle abfcong\triangle cde$. 6. click here to insert. 7. click here to insert. 7. subtraction property. 8. $ab = cd$. 8. click here to insert.

Explanation:

Step1: Define right angles

Since $\overline{AF}\perp\overline{BD}$ and $\overline{CE}\perp\overline{BD}$, $\angle AFB=\angle CED = 90^{\circ}$.

Step2: Prove equal segments

Given $BE = FD$, then $BE + EF=FD+EF$, so $BF = DE$.

Step3: Apply HL criterion

In $\triangle ABF$ and $\triangle CDE$, $AB = CD$ (given) and $BF=DE$ (proven), and $\angle AFB=\angle CED = 90^{\circ}$, so $\triangle ABF\cong\triangle CDE$ by HL.

Answer:

Since $\overline{AF}\perp\overline{BD}$ and $\overline{CE}\perp\overline{BD}$, $\angle AFB=\angle CED = 90^{\circ}$ (Definition of perpendicular lines).
We know that $BE = FD$, so $BE+EF=FD + EF$ (Addition Property), which means $BF=DE$.
We are given that $AB = CD$.
In right - triangles $\triangle ABF$ and $\triangle CDE$, we have:

• Hypotenuse $AB = CD$ (Given)
• Leg $BF=DE$ (Proven above)

By the Hypotenuse - Leg (HL) congruence criterion for right - triangles, $\triangle ABF\cong\triangle CDE$.