Question

1. abc is an isosceles triangle find the measure of ∠dba. find the measure of ∠bcd. find the measure of ∠adc.

Explanation:

Step1: Find the measure of \(\angle DBA\)

Since \(ABC\) is an isosceles triangle and \(BD\) is given with some angles. But from the figure, if we assume \(BD\) is part of a right - angled triangle at \(D\) and given \(\angle BDA = 90^{\circ}\), \(\angle BAD=28^{\circ}\). Using the angle - sum property of a triangle (\(\angle DBA+\angle BAD+\angle BDA = 180^{\circ}\)).

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Step2: Find the measure of \(\angle BCD\)

Since \(ABC\) is isosceles (\(AB = AC\)), and \(BD = AD\) (by some property of the isosceles triangle and the construction of the figure, assume \(BD = AD\) as \(D\) is a point such that \(BD\perp AD\) and \(CD\) is a bisector or by congruency). \(\angle BCD=\angle BAD = 28^{\circ}\) (by congruent triangles \(ABD\) and \(ACD\) ( \(AB = AC\), \(BD = AD\), \(CD = CD\) (common side), SSS congruency)).

Step3: Find the measure of \(\angle ADC\)

Using the angle - sum property of a quadrilateral \(ADBC\) (or by considering triangles \(ABD\) and \(ACD\)). Since \(\angle ADB = 90^{\circ}\) and \(\angle BDC=\angle ADC\) (by congruency of \(\triangle ABD\) and \(\triangle ACD\)). The sum of angles around point \(D\) is \(360^{\circ}\). \(\angle ADB+\angle BDC+\angle ADC+\angle CDB = 360^{\circ}\), but since \(\triangle ABD\cong\triangle ACD\), \(\angle ADC = 135^{\circ}\) (another way: \(\angle BAD = 28^{\circ}\), \(\angle CAD=28^{\circ}\), in quadrilateral \(ADBC\) with \(AB = AC\), \(BD = AD\), \(CD = CD\), using the fact that \(\angle ADB = 90^{\circ}\), and by angle - sum of a quadrilateral \(360^{\circ}\), \(\angle ADC=135^{\circ}\))

Answer:

\(\angle DBA = 62^{\circ}\), \(\angle BCD=28^{\circ}\), \(\angle ADC = 135^{\circ}\)