Question

problem 23
given: (overline{ab}congoverline{bc});
(overline{bm}) is an altitude
of (\triangle abc).
prove: point (m) is the
midpoint of (overline{ac}).

Explanation:

Step1: Identify the right - angled triangles

Since \(BM\) is an altitude of \(\triangle ABC\), \(\angle AMB=\angle CMB = 90^{\circ}\). So, \(\triangle AMB\) and \(\triangle CMB\) are right - angled triangles.

Step2: Use the Hypotenuse - Leg (HL) congruence criterion

In right - angled triangles \(\triangle AMB\) and \(\triangle CMB\), we know that \(\overline{AB}\cong\overline{BC}\) (given) and \(\overline{BM}\cong\overline{BM}\) (common side). By the Hypotenuse - Leg (HL) congruence criterion for right - angled triangles, \(\triangle AMB\cong\triangle CMB\).

Step3: Use the property of congruent triangles

If \(\triangle AMB\cong\triangle CMB\), then their corresponding parts are congruent. So, \(\overline{AM}\cong\overline{CM}\).

Answer:

1. \(\overline{BM}\cong\overline{BM}\); Reason: Reflexive property of congruence
2. Reason: Hypotenuse - Leg (HL) congruence criterion. Since \(M\) is a point on \(AC\) such that \(\overline{AM}\cong\overline{CM}\), point \(M\) is the midpoint of \(\overline{AC}\) (by the definition of a midpoint: a point that divides a segment into two congruent segments).