Question

Triangles ΔABC ≅ ΔBAD so that C and D lie in the opposite semi-planes of segment AB. Prove that segment CD bisects segment AD.

Answer 1

Answer:

See explanation

Step-by-step explanation:

Triangles ΔABC and ΔBAD are congruent. So,

• AB ≅ BA;
• AC ≅ BD;
• BC ≅ AD;
• ∠ABC ≅ ∠BAD;
• ∠BCA ≅ ∠ADB;
• ∠CAB ≅ ∠DBA.

Consider triangles AEC and BED. In these triangles,

• AC ≅ BD;
• ∠EAC ≅ ∠EBD (because ∠CBA ≅ ∠BAD);
• ∠AEC ≅ ∠BED (as vertical angles).

So, ΔAEC ≅ ΔBED. Thus,

AE ≅ EB.

This means that segment CD bisects segment AD.

Answer 2

Congruent triangles are triangles with equal corresponding sides.

See below for the required proof

From the question, we have:

$\mathbf{\triangle ABC \cong \triangle BAD}$

The above means that, the following sides are congruent

$\mathbf{AB \cong BA}$

$\mathbf{AC \cong BD}$

$\mathbf{BC \cong AD}$

Similarly, the following angles are corresponding

$\mathbf{\angle ABC \cong \angle BAD}$

By vertical angle theorem, we have the following corresponding angles:

$\mathbf{\angle AEC \cong \angle BED}$

$\mathbf{\angle EAC \cong \angle EBD}$

Notice that points A and D are common in the above angles

So, we have:

$\mathbf{AE \cong EB}$

Hence,

Line segment CD bisects line segment AD

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