Chapter 3.5, Problem 12PSC

To determine

To prove:One pair of angles is congruent.

Explanation of Solution

Given information Two given line segments are congruent, i.e., $\overline{\text{HO}}\cong \overline{\text{MO}}$

Two other given line segments are congruent, i.e., $\overline{\text{JO}}\cong \overline{\text{KO}}$

  $\begin{array}{l}\overline{\text{HJ}}\text{ is an altitude of }\Delta \text{HJK}\\ \overline{\text{MK}}\text{ is an altitude of }\Delta \text{MKJ}\end{array}$

Formula used Congruency means same shape and same size

Congruent parts of congruent triangle are congruent

SAS criteria of congruency

Altitude means perpendicular on line segment and it makes $\text{90}°$ on line segment

Proof

  $\begin{array}{l}\text{In }\Delta \text{OJH and }\Delta \text{OMK, we have}\\ \angle \text{JOH }\cong \angle \text{MOK}\mathrm{.....}\text{vertical angles}\mathrm{......}\left(i\right)\\ \overline{\text{HO}}\cong \overline{\text{MO}}\mathrm{........}\text{given}\\ \therefore \overline{\text{HO}}=\overline{\text{MO}},\mathrm{......}\text{by definition of congruency}\mathrm{...}\left(ii\right)\\ \overline{\text{JO}}\cong \overline{\text{KO}}\mathrm{........}\text{given}\\ \therefore \overline{\text{JO}}=\overline{\text{KO}},\mathrm{......}\text{by definition of congruency}\mathrm{...}\left(iii\right)\\ \Delta \text{OJH }\cong \Delta \text{OMK, SAS criteria of congruency satisfied}\\ \angle \text{HJO= }\angle \text{MKO, CPCTC}\mathrm{........}\left(a\right)\\ \overline{\text{HJ}}\text{ is an altitude of }\Delta \text{HJK and}\\ \overline{\text{MK}}\text{ is an altitude of }\Delta \text{MKJ}\\ \text{⇒}\angle \text{HJK = }\angle \text{MKJ=90}°\\ \text{⇒}\angle \text{HJO+}\angle \text{1= }\angle \text{MKO+}\angle \text{2}\\ \text{⇒}\angle \text{1= }\angle \text{2}\mathrm{........}\left(\because \text{from }\left(a\right),\angle \text{HJO= }\angle \text{MKO}\right)\\ \text{Hence proved}\end{array}$