Chapter 1.6, Problem 12PSD

To determine

To Prove: $m\angle D=90°+\frac{1}{2}\left(m\angle A\right)$

Explanation of Solution

Given:

  

  $\begin{array}{l}m\angle A+m\angle ABC+m\angle ACB=180°\\ m\angle D+m\angle DBC+m\angle DCB=180°\\ \overrightarrow{BD}\text{bisects }\angle ABC\\ \overrightarrow{CD}\text{bisects }\angle ACB\end{array}$

Concept Used:

Bisector of an angle bisects into two equal parts.

Proof:

Here, we have

  $\begin{array}{l}\because \overrightarrow{BD}\text{bisects }\angle ABC\\ \Rightarrow m\angle ADB=m\angle DBC\\ \overrightarrow{CD}\text{bisects }\angle ACB\\ \Rightarrow m\angle ACD=m\angle DCB\\ \text{In }\Delta ABC,\\ m\angle A+m\angle ABC+m\angle ACB=180°\\ \text{In }\Delta DBC,\\ m\angle D+m\angle DBC+m\angle DCB=180°\\ \Rightarrow m\angle D=180°-m\angle DBC-m\angle DCB\\ \Rightarrow m\angle D=180°-\frac{1}{2}m\angle ABC-\frac{1}{2}m\angle ACB\\ \Rightarrow m\angle D=180°-\frac{1}{2}\left(m\angle ABC+m\angle ACB\right)\\ \Rightarrow m\angle D=180°-\frac{1}{2}\left(180°-m\angle A\right)\\ \Rightarrow m\angle D=180°-90°+\frac{1}{2}\left(m\angle A\right)\\ \Rightarrow m\angle D=90°+\frac{1}{2}\left(m\angle A\right)\end{array}$