Chapter 3.4, Problem 12PSC

To determine

To prove:In a given triangle, Angle 1 is complementary to angle 2.

Explanation of Solution

Given information:

In a given triangle,

One line segment (AC) is the altitude other line segment (BD)

Line segment (AC) is the median

Angle (DAC) is complementary to angle (B)

Formula used:

Equality postulate, which states, In a triangle, median divides its sides in two equal parts.

Two angles are said to be complementary if sum of the angles equal to $\text{90}°$

Sum of the angles in a triangle is $\text{180}°$

Proof:

Since, it is given

AC is altitude

  $\angle \text{ACD=}\angle \text{ACB=90}°$

Also, it is given that

  $\angle \text{BAC is complmentary to }\angle \text{ADC}$

Therefore, if $\angle \text{BAC= x}°$ then $\angle \text{ADC=}\left(\text{90}-\text{x}\right)°$

  $\begin{array}{l}\text{In }\Delta \text{ACD, we have}\\ \angle \text{ADC}+\angle \text{ACD +}\angle \text{CAD=180}°\\ \left(\because \text{sum of the angles in a triangle is 180}°\right)\\ \therefore \left(\text{90}-\text{x}\right)°+\angle \text{CAD + 90}°\text{=180}°\\ \text{⇒}\angle \text{CAD}=\text{x}°\mathrm{........}\left(i\right)\\ \text{In }\Delta \text{ACB, we have}\\ \angle \text{ABC}+\angle \text{ACB +}\angle \text{BAC=180}°\\ \left(\because \text{sum of the angles in a triangle is 180}°\right)\\ \therefore \angle \text{ABC}+\text{90}°\text{ +x}°\text{=180}°\\ \text{⇒}\angle \text{ABC+ x}°=180°-90°\\ \text{⇒}\angle \text{ABC+ }\angle \text{CAD}=90°\mathrm{......}\text{since, from }\left(i\right),\angle \text{CAD}=\text{x}°\end{array}$

Hence, Proved.