Question

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

Explanation:

Step1: Define right - angles

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

Step2: Get equal segments

Given \(BE = FD\), add \(EF\) to both sides: \(BE+EF=FD + EF\), so \(BF = DE\).

Step3: Recall given side

Given \(AB = CD\).

Step4: List congruent parts

In \(\triangle ABF\) and \(\triangle CDE\), we have two sides and the included right - angle (or hypotenuse and leg for right - triangles) congruent.

Step5: Prove congruence

By HL (for right - triangles) or SAS, \(\triangle ABF\cong\triangle CDE\).

Answer:

1. Since \(AF\perp BD\) and \(CE\perp BD\), \(\angle AFB=\angle CED = 90^{\circ}\) (Definition of perpendicular lines).
2. Given \(BE = FD\), then \(BE+EF=FD + EF\) (Addition Property), so \(BF=DE\).
3. Given \(AB = CD\).
4. In \(\triangle ABF\) and \(\triangle CDE\), we have \(AB = CD\), \(\angle AFB=\angle CED\), and \(BF = DE\).
5. So, \(\triangle ABF\cong\triangle CDE\) (by the Hypotenuse - Leg (HL) congruence criterion if considering right - triangles or Side - Angle - Side (SAS) congruence criterion).